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Hyperbola

In mathematics, a hyperbola (pl. hyperbolas or hyperbolae ; adj. hyperbolic ) is a type of smooth curve lying in a plane, defined by its geometric properties or by equations for which it is the solution set. A hyperbola has two pieces, called connected components or branches, that are mirror images of each other and resemble two infinite bows. The hyperbola is one of the three kinds of conic section, formed by the intersection of a plane and a double cone. (The other conic sections are the parabola and the ellipse. A circle is a special case of an ellipse.) If the plane intersects both halves of the double cone but does not pass through the apex of the cones, then the conic is a hyperbola.

Besides being a conic section, a hyperbola can arise as the locus of points whose difference of distances to two fixed foci is constant, as a curve for each point of which the rays to two fixed foci are reflections across the tangent line at that point, or as the solution of certain bivariate quadratic equations such as the reciprocal relationship $$xy = 1.$$ In practical applications, a hyperbola can arise as the path followed by the shadow of the tip of a sundial's gnomon, the shape of an open orbit such as that of a celestial object exceeding the escape velocity of the nearest gravitational body, or the scattering trajectory of a subatomic particle, among others.

Each branch of the hyperbola has two arms which become straighter (lower curvature) further out from the center of the hyperbola. Diagonally opposite arms, one from each branch, tend in the limit to a common line, called the asymptote of those two arms. So there are two asymptotes, whose intersection is at the center of symmetry of the hyperbola, which can be thought of as the mirror point about which each branch reflects to form the other branch. In the case of the curve $$y(x) = 1/x$$ the asymptotes are the two coordinate axes.

Hyperbolas share many of the ellipses' analytical properties such as eccentricity, focus, and directrix. Typically the correspondence can be made with nothing more than a change of sign in some term. Many other mathematical objects have their origin in the hyperbola, such as hyperbolic paraboloids (saddle surfaces), hyperboloids ("wastebaskets"), hyperbolic geometry (Lobachevsky's celebrated non-Euclidean geometry), hyperbolic functions (sinh, cosh, tanh, etc.), and gyrovector spaces (a geometry proposed for use in both relativity and quantum mechanics which is not Euclidean).

Etymology and history
The word "hyperbola" derives from the Greek ὑπερβολή, meaning "over-thrown" or "excessive", from which the English term hyperbole also derives. Hyperbolae were discovered by Menaechmus in his investigations of the problem of doubling the cube, but were then called sections of obtuse cones. The term hyperbola is believed to have been coined by Apollonius of Perga (c. 262) in his definitive work on the conic sections, the Conics. The names of the other two general conic sections, the ellipse and the parabola, derive from the corresponding Greek words for "deficient" and "applied"; all three names are borrowed from earlier Pythagorean terminology which referred to a comparison of the side of rectangles of fixed area with a given line segment. The rectangle could be "applied" to the segment (meaning, have an equal length), be shorter than the segment or exceed the segment.

As locus of points
A hyperbola can be defined geometrically as a set of points (locus of points) in the Euclidean plane:

The midpoint $$M$$ of the line segment joining the foci is called the center of the hyperbola. The line through the foci is called the major axis. It contains the vertices $$V_1, V_2$$, which have distance $$a$$ to the center. The distance $$c$$ of the foci to the center is called the focal distance or linear eccentricity. The quotient $$\tfrac c a$$ is the eccentricity $$e$$.

The equation $$\left|\left|PF_2\right| - \left|PF_1\right|\right| = 2a$$ can be viewed in a different way (see diagram):

If $$c_2$$ is the circle with midpoint $$F_2$$ and radius $$2a$$, then the distance of a point $$P$$ of the right branch to the circle $$c_2$$ equals the distance to the focus $$F_1$$: $$|PF_1|=|Pc_2|.$$ $$c_2$$ is called the circular directrix (related to focus $$F_2$$) of the hyperbola. In order to get the left branch of the hyperbola, one has to use the circular directrix related to $$F_1$$. This property should not be confused with the definition of a hyperbola with help of a directrix (line) below.

Hyperbola with equation $y = A/x$
[[File:Hyperbeln-gs-3.svg|thumb|Three rectangular hyperbolas $$y = A / x$$ with the coordinate axes as asymptotes

red: A = 1; magenta: A = 4; blue: A = 9]] If the xy-coordinate system is rotated about the origin by the angle $$+45^\circ$$ and new coordinates $$\xi,\eta$$ are assigned, then $$x = \tfrac{\xi+\eta}{\sqrt{2}},\; y = \tfrac{-\xi+\eta}{\sqrt{2}} $$.

The rectangular hyperbola $$\tfrac{x^2-y^2}{a^2} = 1$$ (whose semi-axes are equal) has the new equation $$\tfrac{2\xi\eta}{a^2} = 1$$. Solving for $$\eta$$ yields $$\eta = \tfrac{a^2/2}{\xi} \. $$

Thus, in an xy-coordinate system the graph of a function $$f: x \mapsto \tfrac{A}{x},\; A>0\;, $$ with equation $$y = \frac{A}{x}\;, A>0\; ,$$ is a rectangular hyperbola entirely in the first and third quadrants with
 * the coordinate axes as asymptotes,
 * the line $$y = x$$ as major axis ,
 * the center $$(0,0)$$ and the semi-axis $$ a = b = \sqrt{2A} \; ,$$
 * the vertices $$\left(\sqrt{A},\sqrt{A}\right), \left(-\sqrt{A},-\sqrt{A}\right) \; ,$$
 * the semi-latus rectum and radius of curvature  at the vertices $$ p=a=\sqrt{2A} \; ,$$
 * the linear eccentricity $$c=2\sqrt{A}$$ and the eccentricity $$e=\sqrt{2} \; ,$$
 * the tangent $$y=-\tfrac{A}{x_0^2}x+2\tfrac{A}{x_0}$$ at point $$(x_0,A/x_0)\; .$$

A rotation of the original hyperbola by $$-45^\circ$$ results in a rectangular hyperbola entirely in the second and fourth quadrants, with the same asymptotes, center, semi-latus rectum, radius of curvature at the vertices, linear eccentricity, and eccentricity as for the case of $$+45^\circ$$ rotation, with equation $$y = -\frac{A}{x} \;, A>0\; ,$$
 * the semi-axes $$ a = b = \sqrt{2A} \; ,$$
 * the line $$ y = -x$$ as major axis,
 * the vertices $$\left(-\sqrt{A},\sqrt{A}\right), \left(\sqrt{A},-\sqrt{A}\right) \; .$$

Shifting the hyperbola with equation $$y=\frac{A}{x}, \ A\ne 0\ ,$$ so that the new center is $(c_0,d_0)$, yields the new equation $$y=\frac{A}{x-c_0}+d_0\; ,$$ and the new asymptotes are $$x=c_0 $$ and $$y=d_0$$. The shape parameters $$a,b,p,c,e$$ remain unchanged.

By the directrix property
The two lines at distance $d = \frac{a^2}c$ from the center and parallel to the minor axis are called directrices of the hyperbola (see diagram).

For an arbitrary point $$P$$ of the hyperbola the quotient of the distance to one focus and to the corresponding directrix (see diagram) is equal to the eccentricity: $$\frac{|PF_1|}{|Pl_1|} = \frac{|PF_2|}{|Pl_2|} = e= \frac{c}{a} \, .$$ The proof for the pair $$F_1, l_1$$ follows from the fact that $$|PF_1|^2 = (x-c)^2+y^2,\ |Pl_1|^2 = \left(x-\tfrac{a^2}{c}\right)^2$$ and $$y^2 = \tfrac{b^2}{a^2}x^2-b^2$$ satisfy the equation $$|PF_1|^2-\frac{c^2}{a^2}|Pl_1|^2 = 0\ .$$ The second case is proven analogously. The inverse statement is also true and can be used to define a hyperbola (in a manner similar to the definition of a parabola):

For any point $$F$$ (focus), any line $$l$$ (directrix) not through $$F$$ and any real number $$e$$ with $$e > 1$$ the set of points (locus of points), for which the quotient of the distances to the point and to the line is $$e$$ $$H = \left\{P \, \Biggr| \, \frac{|PF|}{|Pl|} = e\right\} $$ is a hyperbola.

(The choice $$e = 1$$ yields a parabola and if $$e < 1$$ an ellipse.)

Proof
Let $$F=(f,0) ,\ e >0$$ and assume $$(0,0)$$ is a point on the curve. The directrix $$l$$ has equation $$x=-\tfrac{f}{e}$$. With $$P=(x,y)$$, the relation $$|PF|^2 = e^2|Pl|^2$$ produces the equations
 * $$(x-f)^2+y^2 = e^2\left(x+\tfrac{f}{e}\right)^2 = (e x+f)^2$$ and $$x^2(e^2-1)+2xf(1+e)-y^2 = 0.$$

The substitution $$p=f(1+e)$$ yields $$x^2(e^2-1)+2px-y^2 = 0.$$ This is the equation of an ellipse ($$e<1$$) or a parabola ($$e=1$$) or a hyperbola ($$e>1$$). All of these non-degenerate conics have, in common, the origin as a vertex (see diagram).

If $$e > 1$$, introduce new parameters $$a,b$$ so that $$e^2-1 = \tfrac{b^2}{a^2}, \text { and }\ p = \tfrac{b^2}{a}$$, and then the equation above becomes $$\frac{(x+a)^2}{a^2} - \frac{y^2}{b^2} = 1 \, ,$$ which is the equation of a hyperbola with center $$(-a,0)$$, the x-axis as major axis and the major/minor semi axis $$a,b$$.

Construction of a directrix
Because of $$c \cdot \tfrac{a^2}{c}=a^2$$ point $$L_1$$ of directrix $$l_1$$ (see diagram) and focus $$F_1$$ are inverse with respect to the circle inversion at circle $$x^2+y^2=a^2$$ (in diagram green). Hence point $$E_1$$ can be constructed using the theorem of Thales (not shown in the diagram). The directrix $$l_1$$ is the perpendicular to line $$\overline{F_1F_2}$$ through point $$E_1$$.

Alternative construction of $$E_1$$: Calculation shows, that point $$E_1$$ is the intersection of the asymptote with its perpendicular through $$F_1$$ (see diagram).

As plane section of a cone
The intersection of an upright double cone by a plane not through the vertex with slope greater than the slope of the lines on the cone is a hyperbola (see diagram: red curve). In order to prove the defining property of a hyperbola (see above) one uses two Dandelin spheres $$d_1, d_2$$, which are spheres that touch the cone along circles $c_1$, $$c_2 $$ and the intersecting (hyperbola) plane at points $$F_1$$ and $F_2$. It turns out: $$F_1, F_2$$ are the foci of the hyperbola.
 * 1) Let  $$P$$ be an arbitrary point of the intersection curve.
 * 2) The generatrix of the cone containing $$P$$ intersects circle $$c_1$$ at point $$A$$ and circle $$c_2$$ at a point $$B$$.
 * 3) The line segments $$\overline{PF_1}$$ and $$\overline{PA}$$ are tangential to the sphere $$d_1$$ and, hence, are of equal length.
 * 4) The line segments $$\overline{PF_2}$$ and $$\overline{PB}$$ are tangential to the sphere $$d_2$$ and, hence, are of equal length.
 * 5) The result is: $$|PF_1| - |PF_2| = |PA| - |PB| = |AB|$$ is independent of the hyperbola point $P$, because no matter where point $$P$$ is, $$A, B$$ have to be on circles $c_1$, $c_2 $, and line segment $$AB$$ has to cross the apex. Therefore, as point $$P$$ moves along the red curve (hyperbola), line segment $$\overline{AB}$$ simply rotates about apex without changing its length.

Pin and string construction
The definition of a hyperbola by its foci and its circular directrices (see above) can be used for drawing an arc of it with help of pins, a string and a ruler:

 * 1) Choose the foci $$F_1,F_2$$, the vertices $$V_1,V_2$$ and one of the circular directrices, for example  $$c_2$$ (circle with radius $$2a$$)
 * 2) A ruler is fixed at point $$F_2$$ free to rotate around $$F_2$$. Point $$B$$ is marked at distance $$2a$$.
 * 3) A string with length $$ |AB|$$ is prepared.
 * 4) One end of the string is pinned at point $$A$$ on the ruler, the other end is pinned to point $$F_1$$.
 * 5) Take a pen and hold the string tight to the edge of the ruler.
 * 6) Rotating the ruler around $$F_2$$ prompts the pen to draw an arc of the right branch of the hyperbola, because of $$|PF_1| = |PB|$$ (see the definition of a hyperbola by circular directrices).

Steiner generation of a hyperbola
The following method to construct single points of a hyperbola relies on the Steiner generation of a non degenerate conic section:

For the generation of points of the hyperbola $$\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2} = 1$$ one uses the pencils at the vertices $$V_1,V_2$$. Let $$P = (x_0,y_0)$$ be a point of the hyperbola and $$A = (a,y_0), B = (x_0,0)$$. The line segment $$\overline{BP}$$ is divided into n equally-spaced segments and this division is projected parallel with the diagonal $$AB$$ as direction onto the line segment $$\overline{AP}$$ (see diagram). The parallel projection is part of the projective mapping between the pencils at $$V_1$$ and $$V_2$$ needed. The intersection points of any two related lines $$S_1 A_i$$ and $$S_2 B_i$$ are points of the uniquely defined hyperbola.

Remarks:
 * The subdivision could be extended beyond the points $$A$$ and $$B$$ in order to get more points, but the determination of the intersection points would become more inaccurate. A better idea is extending the points already constructed by symmetry (see animation).
 * The Steiner generation exists for ellipses and parabolas, too.
 * The Steiner generation is sometimes called a parallelogram method because one can use other points rather than the vertices, which starts with a parallelogram instead of a rectangle.

Inscribed angles for hyperbolas $y = a/(x − b) + c$ and the 3-point-form
A hyperbola with equation $$y=\tfrac{a}{x-b}+c,\ a \ne 0 $$ is uniquely determined by three points $$(x_1,y_1),\;(x_2,y_2),\; (x_3,y_3)$$ with different x- and y-coordinates. A simple way to determine the shape parameters $$a,b,c$$ uses the inscribed angle theorem for hyperbolas:

Analogous to the inscribed angle theorem for circles one gets the

A consequence of the inscribed angle theorem for hyperbolas is the

As an affine image of the unit hyperbola $x^{2} − y^{2} = 1$
Another definition of a hyperbola uses affine transformations:

Parametric representation
An affine transformation of the Euclidean plane has the form $$\vec x \to \vec f_0+A\vec x$$, where $$A$$ is a regular matrix (its determinant is not 0) and $$\vec f_0$$ is an arbitrary vector. If $$\vec f_1, \vec f_2$$ are the column vectors of the matrix $$A$$, the unit hyperbola $$(\pm\cosh(t),\sinh(t)), t \in \R,$$ is mapped onto the hyperbola

$$\vec x = \vec p(t)=\vec f_0 \pm\vec f_1 \cosh t +\vec f_2 \sinh t \ .$$

$$\vec f_0$$ is the center, $$\vec f_0+ \vec f_1$$ a point of the hyperbola and $$\vec f_2$$ a tangent vector at this point.

Vertices
In general the vectors $$\vec f_1, \vec f_2$$ are not perpendicular. That means, in general $$\vec f_0\pm \vec f_1$$ are not the vertices of the hyperbola. But $$\vec f_1\pm \vec f_2$$ point into the directions of the asymptotes. The tangent vector at point $$\vec p(t)$$ is $$\vec p'(t) = \vec f_1\sinh t + \vec f_2\cosh t \ .$$ Because at a vertex the tangent is perpendicular to the major axis of the hyperbola one gets the parameter $$t_0$$ of a vertex from the equation $$\vec p'(t)\cdot \left(\vec p(t) -\vec f_0\right) = \left(\vec f_1\sinh t + \vec f_2\cosh t\right) \cdot \left(\vec f_1 \cosh t +\vec f_2 \sinh t\right) = 0$$ and hence from $$\coth (2t_0)= -\tfrac{\vec f_1^{\, 2}+\vec f_2^{\, 2}}{2\vec f_1 \cdot \vec f_2} \ ,$$ which yields

$$t_0=\tfrac{1}{4}\ln\tfrac{\left(\vec f_1-\vec f_2\right)^2}{\left(\vec f_1+\vec f_2\right)^2}.$$

The formulae $\cosh^2 x + \sinh^2 x = \cosh 2x$, $2\sinh x \cosh x = \sinh 2x$, and $$\operatorname{arcoth} x = \tfrac{1}{2}\ln\tfrac{x+1}{x-1}$$ were used.

The two vertices of the hyperbola are $$\vec f_0\pm\left(\vec f_1\cosh t_0 +\vec f_2 \sinh t_0\right).$$

Implicit representation
Solving the parametric representation for $$ \cosh t, \sinh t$$ by Cramer's rule and using $$\;\cosh^2t-\sinh^2t -1 = 0\; $$, one gets the implicit representation $$\det\left(\vec x\!-\!\vec f\!_0,\vec f\!_2\right)^2 - \det\left(\vec f\!_1,\vec x\!-\!\vec f\!_0\right)^2 - \det\left(\vec f\!_1,\vec f\!_2\right)^2 = 0 .$$

Hyperbola in space
The definition of a hyperbola in this section gives a parametric representation of an arbitrary hyperbola, even in space, if one allows $$\vec f\!_0, \vec f\!_1, \vec f\!_2$$ to be vectors in space.

As an affine image of the hyperbola $y = 1/x$
Because the unit hyperbola $$x^2-y^2=1$$ is affinely equivalent to the hyperbola $$y=1/x$$, an arbitrary hyperbola can be considered as the affine image (see previous section) of the hyperbola $y = 1/x \, $:

$$\vec x = \vec p(t) = \vec f_0 + \vec f_1 t + \vec f_2 \tfrac{1}{t}, \quad t\ne 0\, .$$

$$M: \vec f_0 $$ is the center of the hyperbola, the vectors $$\vec f_1, \vec f_2 $$ have the directions of the asymptotes and $$\vec f_1 + \vec f_2 $$ is a point of the hyperbola. The tangent vector is $$\vec p'(t)=\vec f_1 - \vec f_2 \tfrac{1}{t^2}.$$ At a vertex the tangent is perpendicular to the major axis. Hence $$\vec p'(t)\cdot \left(\vec p(t) -\vec f_0\right) = \left(\vec f_1 - \vec f_2 \tfrac{1}{t^2}\right)\cdot\left(\vec f_1 t+ \vec f_2 \tfrac{1}{t}\right) = \vec f_1^2t-\vec f_2^2 \tfrac{1}{t^3} = 0$$ and the parameter of a vertex is

$$t_0= \pm \sqrt[4]{\frac{\vec f_2^2}{\vec f_1^2}}.$$

$$\left|\vec f\!_1\right| = \left|\vec f\!_2\right|$$ is equivalent to $$t_0 = \pm 1$$ and $$\vec f_0 \pm (\vec f_1+\vec f_2)$$ are the vertices of the hyperbola.

The following properties of a hyperbola are easily proven using the representation of a hyperbola introduced in this section.

Tangent construction
The tangent vector can be rewritten by factorization: $$\vec p'(t)=\tfrac{1}{t}\left(\vec f_1t - \vec f_2 \tfrac{1}{t}\right) \ .$$ This means that

This property provides a way to construct the tangent at a point on the hyperbola.

This property of a hyperbola is an affine version of the 3-point-degeneration of Pascal's theorem.

The area of the grey parallelogram $$MAPB$$ in the above diagram is $$\text{Area} = \left|\det\left( t\vec f_1, \tfrac{1}{t}\vec f_2\right)\right| = \left|\det\left(\vec f_1,\vec f_2\right)\right| = \cdots = \frac{a^2+b^2}{4} $$ and hence independent of point $$P$$. The last equation follows from a calculation for the case, where $$P$$ is a vertex and the hyperbola in its canonical form $$\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1 \, .$$
 * Area of the grey parallelogram:

Point construction

For a hyperbola with parametric representation $$\vec x = \vec p(t) = \vec f_1 t + \vec f_2 \tfrac{1}{t}$$ (for simplicity the center is the origin) the following is true:

The simple proof is a consequence of the equation $$\tfrac{1}{t_1}\vec a = \tfrac{1}{t_2}\vec b$$.

This property provides a possibility to construct points of a hyperbola if the asymptotes and one point are given.

This property of a hyperbola is an affine version of the 4-point-degeneration of Pascal's theorem.

Tangent–asymptotes triangle
For simplicity the center of the hyperbola may be the origin and the vectors $$\vec f_1,\vec f_2$$ have equal length. If the last assumption is not fulfilled one can first apply a parameter transformation (see above) in order to make the assumption true. Hence $$\pm (\vec f_1 + \vec f_2)$$ are the vertices, $$\pm(\vec f_1-\vec f_2)$$ span the minor axis and one gets $$|\vec f_1 + \vec f_2| = a$$ and $$|\vec f_1 - \vec f_2| = b$$.

For the intersection points of the tangent at point $$\vec p(t_0) = \vec f_1 t_0 + \vec f_2 \tfrac{1}{t_0}$$ with the asymptotes one gets the points $$C = 2t_0\vec f_1,\ D = \tfrac{2}{t_0}\vec f_2.$$ The area of the triangle $$M,C,D$$ can be calculated by a 2 × 2 determinant: $$A = \tfrac{1}{2}\Big|\det\left( 2t_0\vec f_1, \tfrac{2}{t_0}\vec f_2\right)\Big| = 2\Big|\det\left(\vec f_1,\vec f_2\right)\Big|$$ (see rules for determinants). $$\left|\det(\vec f_1,\vec f_2)\right|$$ is the area of the rhombus generated by $$\vec f_1,\vec f_2$$. The area of a rhombus is equal to one half of the product of its diagonals. The diagonals are the semi-axes $$a,b$$ of the hyperbola. Hence:

Reciprocation of a circle
The reciprocation of a circle B in a circle C always yields a conic section such as a hyperbola. The process of "reciprocation in a circle C" consists of replacing every line and point in a geometrical figure with their corresponding pole and polar, respectively. The pole of a line is the inversion of its closest point to the circle C, whereas the polar of a point is the converse, namely, a line whose closest point to C is the inversion of the point.

The eccentricity of the conic section obtained by reciprocation is the ratio of the distances between the two circles' centers to the radius r of reciprocation circle C. If B and C represent the points at the centers of the corresponding circles, then

$$e = \frac{\overline{BC}}{r}.$$

Since the eccentricity of a hyperbola is always greater than one, the center B must lie outside of the reciprocating circle C.

This definition implies that the hyperbola is both the locus of the poles of the tangent lines to the circle B, as well as the envelope of the polar lines of the points on B. Conversely, the circle B is the envelope of polars of points on the hyperbola, and the locus of poles of tangent lines to the hyperbola. Two tangent lines to B have no (finite) poles because they pass through the center C of the reciprocation circle C; the polars of the corresponding tangent points on B are the asymptotes of the hyperbola. The two branches of the hyperbola correspond to the two parts of the circle B that are separated by these tangent points.

Quadratic equation
A hyperbola can also be defined as a second-degree equation in the Cartesian coordinates $$(x, y)$$ in the plane,

$$ A_{xx} x^2 + 2 A_{xy} xy + A_{yy} y^2 + 2 B_x x + 2 B_y y + C = 0, $$

provided that the constants $$A_{xx},$$ $$A_{xy},$$ $$A_{yy},$$ $$B_x,$$ $$B_y,$$ and $$C$$ satisfy the determinant condition

$$ D := \begin{vmatrix} A_{xx} & A_{xy} \\ A_{xy} & A_{yy} \end{vmatrix} < 0. $$

This determinant is conventionally called the discriminant of the conic section.

A special case of a hyperbola—the degenerate hyperbola consisting of two intersecting lines—occurs when another determinant is zero:

$$ \Delta := \begin{vmatrix} A_{xx} & A_{xy} & B_x \\ A_{xy} & A_{yy} & B_y \\ B_x & B_y & C \end{vmatrix} = 0. $$

This determinant $$\Delta$$ is sometimes called the discriminant of the conic section.

The general equation's coefficients can be obtained from known semi-major axis $$a,$$ semi-minor axis $$b,$$ center coordinates $$(x_\circ, y_\circ)$$, and rotation angle $$\theta$$ (the angle from the positive horizontal axis to the hyperbola's major axis) using the formulae:

$$\begin{align} A_{xx} &=  -a^2 \sin^2\theta + b^2 \cos^2\theta, & B_{x} &= -A_{xx} x_\circ  -  A_{xy} y_\circ, \\[1ex] A_{yy} &=  -a^2 \cos^2\theta + b^2 \sin^2\theta, & B_{y} &= - A_{xy} x_\circ  - A_{yy} y_\circ, \\[1ex] A_{xy} &= \left(a^2 + b^2\right) \sin\theta \cos\theta, & C &=  A_{xx} x_\circ^2 +  2A_{xy} x_\circ y_\circ + A_{yy} y_\circ^2 - a^2 b^2. \end{align}$$

These expressions can be derived from the canonical equation

$$\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$$

by a translation and rotation of the coordinates $(x, y)$:

$$\begin{alignat}{2} X &= \phantom+\left(x - x_\circ\right) \cos\theta &&+ \left(y - y_\circ\right) \sin\theta, \\ Y &=        -\left(x - x_\circ\right) \sin\theta &&+ \left(y - y_\circ\right) \cos\theta. \end{alignat}$$

Given the above general parametrization of the hyperbola in Cartesian coordinates, the eccentricity can be found using the formula in Conic section.

The center $$(x_c, y_c)$$ of the hyperbola may be determined from the formulae

$$\begin{align} x_c &= -\frac{1}{D} \, \begin{vmatrix} B_x & A_{xy} \\ B_y & A_{yy} \end{vmatrix} \,, \\[1ex] y_c &= -\frac{1}{D} \, \begin{vmatrix} A_{xx} & B_x \\ A_{xy} & B_y \end{vmatrix} \,. \end{align}$$

In terms of new coordinates, $$\xi = x - x_c$$ and $$\eta = y - y_c,$$ the defining equation of the hyperbola can be written

$$ A_{xx} \xi^2 + 2A_{xy} \xi\eta + A_{yy} \eta^2 + \frac \Delta D = 0. $$

The principal axes of the hyperbola make an angle $$\varphi$$ with the positive $$x$$-axis that is given by

$$\tan (2\varphi) = \frac{2A_{xy}}{A_{xx} - A_{yy}}.$$

Rotating the coordinate axes so that the $$x$$-axis is aligned with the transverse axis brings the equation into its canonical form

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$

The major and minor semiaxes $$a$$ and $$b$$ are defined by the equations

$$\begin{align} a^2 &= -\frac{\Delta}{\lambda_1 D} = -\frac{\Delta}{\lambda_1^2 \lambda_2}, \\[1ex] b^2 &= -\frac{\Delta}{\lambda_2 D} = -\frac{\Delta}{\lambda_1 \lambda_2^2}, \end{align}$$

where $$\lambda_1$$ and $$\lambda_2$$ are the roots of the quadratic equation

$$\lambda^2 - \left( A_{xx} + A_{yy} \right)\lambda + D = 0.$$

For comparison, the corresponding equation for a degenerate hyperbola (consisting of two intersecting lines) is

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0.$$

The tangent line to a given point $$(x_0, y_0)$$ on the hyperbola is defined by the equation

$$E x + F y + G = 0$$

where $$E,$$ $$F,$$ and $$G$$ are defined by

$$\begin{align} E &= A_{xx} x_0 + A_{xy} y_0 + B_x, \\[1ex] F &= A_{xy} x_0 + A_{yy} y_0 + B_y, \\[1ex] G &= B_x x_0 + B_y y_0 + C. \end{align}$$

The normal line to the hyperbola at the same point is given by the equation

$$F(x - x_0) - E(y - y_0) = 0.$$

The normal line is perpendicular to the tangent line, and both pass through the same point $$(x_0, y_0).$$

From the equation

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \qquad 0 < b \leq a,$$

the left focus is $$(-ae,0)$$ and the right focus is $$(ae,0), $$ where $$e$$ is the eccentricity. Denote the distances from a point $$(x, y)$$ to the left and right foci as $$r_1$$ and $$r_2.$$ For a point on the right branch,

$$ r_1 - r_2 = 2 a, $$

and for a point on the left branch,

$$ r_2 - r_1 = 2 a. $$

This can be proved as follows:

If $$(x, y)$$ is a point on the hyperbola the distance to the left focal point is

$$ r_1^2 = (x+a e)^2 + y^2 = x^2 + 2 x a e + a^2 e^2 + \left(x^2-a^2\right) \left(e^2-1\right) = (e x + a)^2. $$

To the right focal point the distance is

$$ r_2^2 = (x-a e)^2 + y^2 = x^2 - 2 x a e + a^2 e^2 + \left(x^2-a^2\right) \left(e^2-1\right) = (e x - a)^2. $$

If $$(x, y)$$ is a point on the right branch of the hyperbola then $$ex > a$$ and

$$\begin{align} r_1 &= e x + a, \\ r_2 &= e x - a. \end{align}$$

Subtracting these equations one gets

$$r_1 - r_2 = 2a.$$

If $$(x, y)$$ is a point on the left branch of the hyperbola then $$ex < -a$$ and

$$\begin{align} r_1 &= - e x - a, \\ r_2 &= - e x + a. \end{align}$$

Subtracting these equations one gets

$$r_2 - r_1 = 2a.$$

Equation
If Cartesian coordinates are introduced such that the origin is the center of the hyperbola and the x-axis is the major axis, then the hyperbola is called east-west-opening and
 * the foci are the points $$F_1=(c,0),\ F_2=(-c,0)$$,
 * the vertices are $$V_1=(a, 0),\ V_2=(-a,0)$$.

For an arbitrary point $$(x,y)$$ the distance to the focus $$(c,0)$$ is $\sqrt{ (x-c)^2 + y^2 }$ and to the second focus $\sqrt{ (x+c)^2 + y^2 }$. Hence the point $$(x,y)$$ is on the hyperbola if the following condition is fulfilled $$\sqrt{(x-c)^2 + y^2} - \sqrt{(x+c)^2 + y^2} = \pm 2a \ .$$ Remove the square roots by suitable squarings and use the relation $$b^2 = c^2-a^2$$ to obtain the equation of the hyperbola:

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \ .$$

This equation is called the canonical form of a hyperbola, because any hyperbola, regardless of its orientation relative to the Cartesian axes and regardless of the location of its center, can be transformed to this form by a change of variables, giving a hyperbola that is congruent to the original (see below).

The axes of symmetry or principal axes are the transverse axis (containing the segment of length 2a with endpoints at the vertices) and the conjugate axis (containing the segment of length 2b perpendicular to the transverse axis and with midpoint at the hyperbola's center). As opposed to an ellipse, a hyperbola has only two vertices: $$(a,0),\; (-a,0)$$. The two points $$(0,b),\; (0,-b)$$ on the conjugate axes are not on the hyperbola.

It follows from the equation that the hyperbola is symmetric with respect to both of the coordinate axes and hence symmetric with respect to the origin.

Eccentricity
For a hyperbola in the above canonical form, the eccentricity is given by

$$e=\sqrt{1+\frac{b^2}{a^2}}.$$

Two hyperbolas are geometrically similar to each other – meaning that they have the same shape, so that one can be transformed into the other by rigid left and right movements, rotation, taking a mirror image, and scaling (magnification) – if and only if they have the same eccentricity.

Asymptotes
Solving the equation (above) of the hyperbola for $$y$$ yields $$y=\pm\frac{b}{a} \sqrt{x^2-a^2}.$$ It follows from this that the hyperbola approaches the two lines $$y=\pm \frac{b}{a}x $$ for large values of $$|x|$$. These two lines intersect at the center (origin) and are called asymptotes of the hyperbola $$\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}= 1 \ .$$

With the help of the second figure one can see that
 * $${\color{blue}{(1)}}$$ The perpendicular distance from a focus to either asymptote is $$b$$ (the semi-minor axis).

From the Hesse normal form $$\tfrac{bx\pm ay}{\sqrt{a^2+b^2}}=0 $$ of the asymptotes and the equation of the hyperbola one gets:
 * $${\color{magenta}{(2)}}$$ The product of the distances from a point on the hyperbola to both the asymptotes is the constant $$\tfrac{a^2b^2}{a^2+b^2}\, $$ which can also be written in terms of the eccentricity e as $$\left( \tfrac{b}{e}\right) ^2.$$

From the equation $$y=\pm\frac{b}{a}\sqrt{x^2-a^2}$$ of the hyperbola (above) one can derive:
 * $${\color{green}{(3)}}$$ The product of the slopes of lines from a point P to the two vertices is the constant $$b^2/a^2\ .$$

In addition, from (2) above it can be shown that
 * $${\color{red}{(4)}}$$ The product of the distances from a point on the hyperbola to the asymptotes along lines parallel to the asymptotes is the constant $$\tfrac{a^2+b^2}{4}.$$

Semi-latus rectum
The length of the chord through one of the foci, perpendicular to the major axis of the hyperbola, is called the latus rectum. One half of it is the semi-latus rectum $$p$$. A calculation shows $$p = \frac{b^2}a.$$ The semi-latus rectum $$p$$ may also be viewed as the radius of curvature  at the vertices.

Tangent
The simplest way to determine the equation of the tangent at a point $$(x_0,y_0)$$ is to implicitly differentiate the equation $$\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}= 1$$ of the hyperbola. Denoting dy/dx as y′, this produces $$\frac{2x}{a^2}-\frac{2yy'}{b^2}= 0 \ \Rightarrow \ y'=\frac{x}{y}\frac{b^2}{a^2}\ \Rightarrow \ y=\frac{x_0}{y_0}\frac{b^2}{a^2}(x-x_0) +y_0.$$ With respect to $$\tfrac{x_0^2}{a^2}-\tfrac{y_0^2}{b^2}= 1$$, the equation of the tangent at point $$(x_0,y_0)$$ is $$\frac{x_0}{a^2}x-\frac{y_0}{b^2}y = 1.$$

A particular tangent line distinguishes the hyperbola from the other conic sections. Let f be the distance from the vertex V (on both the hyperbola and its axis through the two foci) to the nearer focus. Then the distance, along a line perpendicular to that axis, from that focus to a point P on the hyperbola is greater than 2f. The tangent to the hyperbola at P intersects that axis at point Q at an angle ∠PQV of greater than 45°.

Rectangular hyperbola
In the case $$a = b$$ the hyperbola is called rectangular (or equilateral), because its asymptotes intersect at right angles. For this case, the linear eccentricity is $$c=\sqrt{2}a$$, the eccentricity $$e=\sqrt{2}$$ and the semi-latus rectum $$p=a$$. The graph of the equation $$y=1/x$$ is a rectangular hyperbola.

Parametric representation with hyperbolic sine/cosine
Using the hyperbolic sine and cosine functions $$\cosh,\sinh$$, a parametric representation of the hyperbola $$\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}= 1$$ can be obtained, which is similar to the parametric representation of an ellipse: $$(\pm a \cosh t, b \sinh t),\, t \in \R \ ,$$ which satisfies the Cartesian equation because $$\cosh^2 t -\sinh^2 t =1 .$$

Further parametric representations are given in the section Parametric equations below.

Conjugate hyperbola
Exchange $$\frac{x^2}{a^2}$$ and $$\frac{y^2}{b^2}$$ to obtain the equation of the conjugate hyperbola (see diagram): $$\frac{y^2}{b^2}-\frac{x^2}{a^2}= 1 \ ,$$ also written as $$\frac{x^2}{a^2}-\frac{y^2}{b^2}= -1 \ .$$

A hyperbola and its conjugate may have diameters which are conjugate. In the theory of special relativity, such diameters may represent axes of time and space, where one hyperbola represents events at a given spatial distance from the center, and the other represents events at a corresponding temporal distance from the center.

Origin at the focus
The polar coordinates used most commonly for the hyperbola are defined relative to the Cartesian coordinate system that has its origin in a focus and its x-axis pointing towards the origin of the "canonical coordinate system" as illustrated in the first diagram.

In this case the angle $$\varphi$$ is called true anomaly.

Relative to this coordinate system one has that

$$r = \frac{p}{1 \mp e \cos \varphi}, \quad p = \frac{b^2}{a}$$

and

$$-\arccos \left(-\frac 1 e\right) < \varphi < \arccos \left(-\frac 1 e\right). $$

Origin at the center
With polar coordinates relative to the "canonical coordinate system" (see second diagram) one has that

$$r =\frac{b}{\sqrt{e^2 \cos^2 \varphi -1}} .\,$$

For the right branch of the hyperbola the range of $$ \varphi $$ is $$-\arccos \left(\frac 1 e\right) < \varphi < \arccos \left(\frac 1 e\right).$$

Parametric equations
A hyperbola with equation $$\tfrac{x^2}{a^2} - \tfrac{y^2}{b^2} = 1$$ can be described by several parametric equations: \begin{cases} x = \pm a \cosh t, \\ y = b \sinh t, \end{cases} \qquad t \in \R. $$ \begin{cases} x = \pm a \dfrac{t^2 + 1}{2t}, \\[1ex] y = b \dfrac{t^2 - 1}{2t}, \end{cases} \qquad t > 0$$ \begin{cases} x = \frac{a}{\cos t} = a \sec t, \\ y = \pm b \tan t, \end{cases} \qquad 0 \le t < 2\pi,\ t \ne \frac{\pi}{2},\ t \ne \frac{3}{2} \pi.$$
 * 1) Through hyperbolic trigonometric functions $$
 * 1) As a rational representation $$
 * 1) Through circular trigonometric functions $$
 * 1) With the tangent slope as parameter:  A parametric representation, which uses the slope $$m$$ of the tangent at a point of the hyperbola can be obtained analogously to the ellipse case: Replace in the ellipse case $$b^2$$ by $$-b^2$$ and use formulae for the hyperbolic functions. One gets $$\vec c_\pm(m) = \left(-\frac{ma^2}{\pm\sqrt{m^2a^2 - b^2}}, \frac{-b^2}{\pm\sqrt{m^2a^2 - b^2}}\right),\quad |m| > b/a.$$ Here, $$\vec c_-$$ is the upper, and $$\vec c_+$$ the lower half of the hyperbola. The points with vertical tangents (vertices $$(\pm a, 0)$$) are not covered by the representation.  The equation of the tangent at point $$\vec c_\pm(m)$$ is $$y = m x \pm\sqrt{m^2a^2 - b^2}.$$ This description of the tangents of a hyperbola is an essential tool for the determination of the orthoptic of a hyperbola.

Hyperbolic functions

Just as the trigonometric functions are defined in terms of the unit circle, so also the hyperbolic functions are defined in terms of the unit hyperbola, as shown in this diagram. In a unit circle, the angle (in radians) is equal to twice the area of the circular sector which that angle subtends. The analogous hyperbolic angle is likewise defined as twice the area of a hyperbolic sector.

Let $$a$$ be twice the area between the $$x$$ axis and a ray through the origin intersecting the unit hyperbola, and define $(x,y) = (\cosh a,\sinh a) = (x, \sqrt{x^2-1})$ as the coordinates of the intersection point. Then the area of the hyperbolic sector is the area of the triangle minus the curved region past the vertex at $$(1,0)$$: $$\begin{align} \frac{a}{2} &= \frac{xy}{2} - \int_1^x \sqrt{t^{2}-1} \, dt \\[1ex] &= \frac{1}{2} \left(x\sqrt{x^2-1}\right) - \frac{1}{2} \left(x\sqrt{x^2-1} - \ln \left(x+\sqrt{x^2-1}\right)\right), \end{align}$$ which simplifies to the area hyperbolic cosine $$a=\operatorname{arcosh}x=\ln \left(x+\sqrt{x^2-1}\right).$$ Solving for $$x$$ yields the exponential form of the hyperbolic cosine: $$x=\cosh a=\frac{e^a+e^{-a}}{2}.$$ From $$x^2-y^2=1$$ one gets $$y=\sinh a=\sqrt{\cosh^2 a - 1}=\frac{e^a-e^{-a}}{2},$$ and its inverse the area hyperbolic sine: $$a=\operatorname{arsinh}y=\ln \left(y+\sqrt{y^2+1}\right).$$ Other hyperbolic functions are defined according to the hyperbolic cosine and hyperbolic sine, so for example $$\operatorname{tanh}a=\frac{\sinh a}{\cosh a}=\frac{e^{2a}-1}{e^{2a}+1}.$$

Reflection property
The tangent at a point $$P$$ bisects the angle between the lines $$\overline{PF_1}, \overline{PF_2}.$$ This is called the optical property or reflection property of a hyperbola. Let $$L$$ be the point on the line $$\overline{PF_2}$$ with the distance $$2a$$ to the focus $$F_2$$ (see diagram, $$a$$ is the semi major axis of the hyperbola). Line $$w$$ is the bisector of the angle between the lines $$\overline{PF_1}, \overline{PF_2}$$. In order to prove that $$w$$ is the tangent line at point $$P$$, one checks that any point $$Q$$ on line $$w$$ which is different from $$P$$ cannot be on the hyperbola. Hence $$w$$ has only point $$P$$ in common with the hyperbola and is, therefore, the tangent at point $$P$$.
 * Proof:

From the diagram and the triangle inequality one recognizes that $$|QF_2|<|LF_2|+|QL|=2a+|QF_1|$$ holds, which means: $$|QF_2|-|QF_1|<2a$$. But if $$Q$$ is a point of the hyperbola, the difference should be $$2a$$.

Midpoints of parallel chords
The midpoints of parallel chords of a hyperbola lie on a line through the center (see diagram).

The points of any chord may lie on different branches of the hyperbola.

The proof of the property on midpoints is best done for the hyperbola $$y=1/x$$. Because any hyperbola is an affine image of the hyperbola $$y=1/x$$ (see section below) and an affine transformation preserves parallelism and midpoints of line segments, the property is true for all hyperbolas:

For two points $$P=\left(x_1,\tfrac {1 }{x_1}\right), \ Q=\left(x_2,\tfrac {1 }{x_2}\right)$$ of the hyperbola $$y=1/x$$

 * the midpoint of the chord is $$M=\left(\tfrac{x_1+x_2}{2},\cdots\right)=\cdots =\tfrac{x_1+x_2}{2}\; \left(1,\tfrac{1}{x_1x_2}\right) \ ;$$

 * the slope of the chord is $$\frac{\tfrac {1 }{x_2}-\tfrac {1 }{x_1}}{x_2-x_1}=\cdots =-\tfrac{1}{x_1x_2} \ .$$

For parallel chords the slope is constant and the midpoints of the parallel chords lie on the line $$y=\tfrac{1}{x_1x_2} \; x \ .$$

Consequence: for any pair of points $$P,Q$$ of a chord there exists a skew reflection with an axis (set of fixed points) passing through the center of the hyperbola, which exchanges the points $$P,Q$$ and leaves the hyperbola (as a whole) fixed. A skew reflection is a generalization of an ordinary reflection across a line $$m$$, where all point-image pairs are on a line perpendicular to $$m$$.

Because a skew reflection leaves the hyperbola fixed, the pair of asymptotes is fixed, too. Hence the midpoint $$M$$ of a chord $$P Q$$ divides the related line segment $$\overline P \, \overline Q$$ between the asymptotes into halves, too. This means that $$|P\overline P|=|Q\overline Q|$$. This property can be used for the construction of further points $$Q$$ of the hyperbola if a point $$P$$ and the asymptotes are given.

If the chord degenerates into a tangent, then the touching point divides the line segment between the asymptotes in two halves.

Orthogonal tangents – orthoptic

For a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \, a>b$ the intersection points of orthogonal tangents lie on the circle $$x^2+y^2=a^2-b^2$$.

This circle is called the orthoptic of the given hyperbola.

The tangents may belong to points on different branches of the hyperbola.

In case of $$a\le b$$ there are no pairs of orthogonal tangents.

Pole-polar relation for a hyperbola
Any hyperbola can be described in a suitable coordinate system by an equation $$\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}= 1$$. The equation of the tangent at a point $$P_0=(x_0,y_0)$$ of the hyperbola is $$\tfrac{x_0x}{a^2}-\tfrac{y_0y}{b^2}=1.$$ If one allows  point $$P_0=(x_0,y_0)$$ to be an arbitrary point different from the origin, then

 * point $$P_0=(x_0,y_0)\ne(0,0)$$ is mapped onto the line $$\frac{x_0x}{a^2}-\frac{y_0y}{b^2}=1 $$, not through the center of the hyperbola.

This relation between points and lines is a bijection.

The inverse function maps

 * line $$y=mx+d,\ d\ne 0$$ onto the point $$\left(-\frac{ma^2}{d},-\frac{b^2}{d}\right)$$ and

 * line $$x=c,\ c\ne 0$$ onto the point $$\left(\frac{a^2}{c},0\right)\ .$$

Such a relation between points and lines generated by a conic is called pole-polar relation or just polarity. The pole is the point, the polar the line. See Pole and polar.

By calculation one checks the following properties of the pole-polar relation of the hyperbola:
 * For a point (pole) on the hyperbola the polar is the tangent at this point (see diagram: $$P_1,\ p_1$$).
 * For a pole $$P$$ outside the hyperbola the intersection points of its polar with the hyperbola are the tangency points of the two tangents passing $$P$$ (see diagram: $$P_2,\ p_2,\ P_3,\ p_3$$).
 * For a point within the hyperbola the polar has no point with the hyperbola in common. (see diagram: $$P_4,\ p_4$$).

Remarks: Pole-polar relations exist for ellipses and parabolas, too.
 * 1) The intersection point of two polars (for example: $$p_2,p_3$$) is the pole of the line through their poles (here: $$P_2,P_3$$).
 * 2) The foci $$(c,0),$$ and $$ (-c,0)$$ respectively and the directrices $$x=\tfrac{a^2}{c}$$ and $$x=-\tfrac{a^2}{c}$$ respectively belong to pairs of  pole and polar.

Other properties

 * The following are concurrent: (1) a circle passing through the hyperbola's foci and centered at the hyperbola's center; (2) either of the lines that are tangent to the hyperbola at the vertices; and (3) either of the asymptotes of the hyperbola.
 * The following are also concurrent: (1) the circle that is centered at the hyperbola's center and that passes through the hyperbola's vertices; (2) either directrix; and (3) either of the asymptotes.
 * Since both the transverse axis and the conjugate axis are axes of symmetry, the symmetry group of a hyperbola is the Klein four-group.
 * The rectangular hyperbolas xy = constant admit group actions by squeeze mappings which have the hyperbolas as invariant sets.

Arc length
The arc length of a hyperbola does not have an elementary expression. The upper half of a hyperbola can be parameterized as

$$y = b\sqrt{\frac{x^{2}}{a^{2}}-1}.$$

Then the integral giving the arc length $$s$$ from $$x_{1}$$ to $$x_{2}$$ can be computed as:

$$s = b\int_{\operatorname{arcosh}\frac{x_{1}}{a}}^{\operatorname{arcosh}\frac{x_{2}}{a}} \sqrt{1+\left(1+\frac{a^{2}}{b^{2}}\right) \sinh ^{2}v} \, \mathrm dv.$$

After using the substitution $$z = iv$$, this can also be represented using the incomplete elliptic integral of the second kind $$E$$ with parameter $$m = k^2$$:

$$s = ib \Biggr[E\left(iv \, \Biggr| \, 1 + \frac{a^2}{b^2}\right)\Biggr]^{\operatorname{arcosh}\frac{x_1}{a}}_{\operatorname{arcosh}\frac{x_2}{a}}.$$

Using only real numbers, this becomes

$$s=b\left[F\left(\operatorname{gd}v\,\Biggr|-\frac{a^2}{b^2}\right) - E\left(\operatorname{gd}v\,\Biggr|-\frac{a^2}{b^2}\right) + \sqrt{1+\frac{a^2}{b^2}\tanh^2 v}\,\sinh v\right]_{\operatorname{arcosh}\tfrac{x_1}{a}}^{\operatorname{arcosh}\tfrac{x_2}{a}}$$

where $$F$$ is the incomplete elliptic integral of the first kind with parameter $$m = k^2$$ and $$\operatorname{gd}v=\arctan\sinh v$$ is the Gudermannian function.

Derived curves
Several other curves can be derived from the hyperbola by inversion, the so-called inverse curves of the hyperbola. If the center of inversion is chosen as the hyperbola's own center, the inverse curve is the lemniscate of Bernoulli; the lemniscate is also the envelope of circles centered on a rectangular hyperbola and passing through the origin. If the center of inversion is chosen at a focus or a vertex of the hyperbola, the resulting inverse curves are a limaçon or a strophoid, respectively.

Elliptic coordinates
A family of confocal hyperbolas is the basis of the system of elliptic coordinates in two dimensions. These hyperbolas are described by the equation

$$ \left(\frac x {c \cos\theta}\right)^2 - \left(\frac y {c \sin\theta}\right)^2 = 1 $$

where the foci are located at a distance c from the origin on the x-axis, and where θ is the angle of the asymptotes with the x-axis. Every hyperbola in this family is orthogonal to every ellipse that shares the same foci. This orthogonality may be shown by a conformal map of the Cartesian coordinate system w = z + 1/z, where z= x + iy are the original Cartesian coordinates, and w=u + iv are those after the transformation.

Other orthogonal two-dimensional coordinate systems involving hyperbolas may be obtained by other conformal mappings. For example, the mapping w = z2 transforms the Cartesian coordinate system into two families of orthogonal hyperbolas.

Conic section analysis of the hyperbolic appearance of circles
Zp-Kugel-Augp-innen.svg of circles on a sphere: The center O of projection is inside the sphere, the image plane is red.

As images of the circles one gets a circle (magenta), ellipses, hyperbolas and lines. The special case of a parabola does not appear in this example.

(If center O were on the sphere, all images of the circles would be circles or lines; see stereographic projection).]] Besides providing a uniform description of circles, ellipses, parabolas, and hyperbolas, conic sections can also be understood as a natural model of the geometry of perspective in the case where the scene being viewed consists of circles, or more generally an ellipse. The viewer is typically a camera or the human eye and the image of the scene a central projection onto an image plane, that is, all projection rays pass a fixed point O, the center. The lens plane is a plane parallel to the image plane at the lens O.

The image of a circle c is 1. a circle, if circle c is in a special position, for example parallel to the image plane and others (see stereographic projection),

2. an ellipse, if c has no point with the lens plane in common,

3. a parabola, if c has one point with the lens plane in common and

4. a hyperbola, if c has two points with the lens plane in common. (Special positions where the circle plane contains point O are omitted.)

These results can be understood if one recognizes that the projection process can be seen in two steps: 1) circle c and point O generate a cone which is 2) cut by the image plane, in order to generate the image.

One sees a hyperbola whenever catching sight of a portion of a circle cut by one's lens plane. The inability to see very much of the arms of the visible branch, combined with the complete absence of the second branch, makes it virtually impossible for the human visual system to recognize the connection with hyperbolas.

Sundials
Hyperbolas may be seen in many sundials. On any given day, the sun revolves in a circle on the celestial sphere, and its rays striking the point on a sundial traces out a cone of light. The intersection of this cone with the horizontal plane of the ground forms a conic section. At most populated latitudes and at most times of the year, this conic section is a hyperbola. In practical terms, the shadow of the tip of a pole traces out a hyperbola on the ground over the course of a day (this path is called the declination line). The shape of this hyperbola varies with the geographical latitude and with the time of the year, since those factors affect the cone of the sun's rays relative to the horizon. The collection of such hyperbolas for a whole year at a given location was called a pelekinon by the Greeks, since it resembles a double-bladed axe.

Multilateration
A hyperbola is the basis for solving multilateration problems, the task of locating a point from the differences in its distances to given points — or, equivalently, the difference in arrival times of synchronized signals between the point and the given points. Such problems are important in navigation, particularly on water; a ship can locate its position from the difference in arrival times of signals from a LORAN or GPS transmitters. Conversely, a homing beacon or any transmitter can be located by comparing the arrival times of its signals at two separate receiving stations; such techniques may be used to track objects and people. In particular, the set of possible positions of a point that has a distance difference of 2a from two given points is a hyperbola of vertex separation 2a whose foci are the two given points.

Path followed by a particle
The path followed by any particle in the classical Kepler problem is a conic section. In particular, if the total energy E of the particle is greater than zero (that is, if the particle is unbound), the path of such a particle is a hyperbola. This property is useful in studying atomic and sub-atomic forces by scattering high-energy particles; for example, the Rutherford experiment demonstrated the existence of an atomic nucleus by examining the scattering of alpha particles from gold atoms. If the short-range nuclear interactions are ignored, the atomic nucleus and the alpha particle interact only by a repulsive Coulomb force, which satisfies the inverse square law requirement for a Kepler problem.

Korteweg–de Vries equation
The hyperbolic trig function $$\operatorname{sech}\, x$$ appears as one solution to the Korteweg–de Vries equation which describes the motion of a soliton wave in a canal.

Angle trisection

As shown first by Apollonius of Perga, a hyperbola can be used to trisect any angle, a well studied problem of geometry. Given an angle, first draw a circle centered at its vertex O, which intersects the sides of the angle at points A and B. Next draw the line segment with endpoints A and B and its perpendicular bisector $$\ell$$. Construct a hyperbola of eccentricity e=2 with $$\ell$$ as directrix and B as a focus. Let P be the intersection (upper) of the hyperbola with the circle. Angle POB trisects angle AOB.

To prove this, reflect the line segment OP about the line $$\ell$$ obtaining the point P' as the image of P. Segment AP' has the same length as segment BP due to the reflection, while segment PP' has the same length as segment BP due to the eccentricity of the hyperbola. As OA, OP', OP and OB are all radii of the same circle (and so, have the same length), the triangles OAP', OPP' and OPB are all congruent. Therefore, the angle has been trisected, since 3×POB = AOB.

Efficient portfolio frontier
In portfolio theory, the locus of mean-variance efficient portfolios (called the efficient frontier) is the upper half of the east-opening branch of a hyperbola drawn with the portfolio return's standard deviation plotted horizontally and its expected value plotted vertically; according to this theory, all rational investors would choose a portfolio characterized by some point on this locus.

Biochemistry
In biochemistry and pharmacology, the Hill equation and Hill-Langmuir equation respectively describe biological responses and the formation of protein–ligand complexes as functions of ligand concentration. They are both rectangular hyperbolae.

Hyperbolas as plane sections of quadrics
Hyperbolas appear as plane sections of the following quadrics:
 * Elliptic cone
 * Hyperbolic cylinder
 * Hyperbolic paraboloid
 * Hyperboloid of one sheet
 * Hyperboloid of two sheets

Other conic sections

 * Circle
 * Ellipse
 * Parabola
 * Degenerate conic

Other related topics

 * Elliptic coordinates, an orthogonal coordinate system based on families of ellipses and hyperbolas.
 * Hyperbolic growth
 * Hyperbolic partial differential equation
 * Hyperbolic sector
 * Hyperboloid structure
 * Hyperbolic trajectory
 * Hyperboloid
 * Multilateration
 * Rotation of axes
 * Translation of axes
 * Unit hyperbola

title: Hyperbola


text:
Talk:HTML

HTML 1.0 was released in 1993 not in 1991
Everyone is saying HTML 1.0 was released in 1991 but actually in 1991 Tim Berners-Lee took on the role of creator for HTML 1.0 and produced a program called WorldWideWeb in 1991 that later changed to Nexus. 1993 marked the introduction of HTML 1.0. At that point, only a few developers have been working on building the website Here is the source that verifies my claim https://spankingsolutions.com/history-of-html/ Spanking710 (talk) 05:53, 22 February 2024 (UTC)

Windows 11 and HTMLLock
And all other things that are in my mind and the upstream documentation is a setup process. At the same time I was just curious about the window 10-5 and how it could work with the new content 41.122.11.57 (talk) 12:54, 25 March 2024 (UTC)

Got vandalized
The article got vandalized with information about some rapper 68.146.42.107 (talk) 09:28, 28 April 2024 (UTC)

Semi-protected edit request on 13 June 2024
"Sorry, HTML is mostly consufed as a markup language, even if the acronym have "Markup Language" is classified as a programming language."

-Sam McGrimer, Official CERN team. ThatMintyLad (talk) 11:12, 13 June 2024 (UTC)
 * Red question icon with gradient background.svg Not done: it's not clear what changes you want to be made. Please mention the specific changes in a "change X to Y" format and provide a reliable source if appropriate.   [[User:CanonNi ]]  (talk • contribs) 11:17, 13 June 2024 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Talk:HTML


text:
Talk:Homeschooling

Biased?
What do you think about moving "According to Elizabeth Bartholet, surveys of homeschoolers show that a majority of homeschoolers in the USA are motivated by "conservative Christian beliefs, and seek to remove their children from mainstream culture". " with updated sourcing to criticism? E. Bartholet is a vocal critic of homeschooling and her writings maybe considered biased.Barbie1979 (talk) 05:59, 1 January 2022 (UTC)

 * The entire article otherwise has to have been written by people willing to extrapolate very limited research. A lot of the referenced being used are from advocacy groups, or corporations that have learning tools to sell. Critics can't be dismissed simply by them having a critical opinion, that's already understood. Compare it to homeopathy maybe, there's many critical voices on the practice, yet they are taken fully seriously on Wiki. Despite being biased. You could start by demonstrating why you think that her writings aren't good, and not just that she disagrees with you.  Kameloso (talk) 01:16, 2 May 2022 (UTC)

"Not educating children at all"
Pinging and. Ft109, I appreciate your efforts, but I'm more inclined to agree with Tacyarg. However, whenever there's disagreements about things, starting a discussion on the revelant talk page should happen instead of repeatedly restoring the content in question. So let's all talk about this and try to come to some sort of understanding?

Personally, I think it's relevant to include information about different types of homeschooling instead of just generalizing across the board. Some people recieve a relatively good education at home, like one of my parents, and others really don't. I also agree about Tacyarg's argument about sourcing. I see where you're coming from in regards to that. Educational neglect does seem to be a term that is used elsewhere. Clovermoss 🍀 (talk) 11:54, 30 November 2022 (UTC)

 * Aspects of the danger of potential educational neglect are already mentioned in the last paragraph of the lead section. However, "not educating children at all" is not a form of homeschooling, but simply the abscence of getting an education (which can also happen to school children who can't cope with the school system). So no, there's nothing to discuss about that; this edit was simply inappropriate, unneutral and factually wrong.-- Maxeto0910 (talk) 23:15, 1 December 2022 (UTC)
 * I disagree that there's nothing to discuss. I'm not even involved in any of this and I saw the content added and removed a few times. I was trying to friendly reach out to the editors in question so we could try to reach some sort of consensus. Clovermoss 🍀  (talk) 08:50, 2 December 2022 (UTC)
 * @Clovermoss As the person who originally added "not educating children at all", I'll admit I should've worded it differently.
 * That being said, I think that it makes sense to acknowledge that educational neglect exists, and not over-generalize. I don't see how only including the views of homeschool-avocate parents helps the article remain neutral.
 * I have since changed it to say "-in a less strict manor, which in some cases may include the children not being educated at all". I think that this makes it more clear that it only applies to some children, and not all. Shane04040404 (talk) 08:35, 3 December 2022 (UTC)
 * One potential is that one might think that educational neglect overlaps with Unschooling and that they're synonymous terms, but I think that would be more of an opinion to combine the two and not a fact, so I think it's appropriate to list them as too separate things. Shane04040404 (talk) 08:43, 3 December 2022 (UTC)
 * I know I haven't edited or commented before now but I've been watching this for a few days, and I have a question; is there a better source than this for the content in question? If that's the best we can provide then it probably doesn't belong, since that's hardly an unbiased review form a site whose advocacy is evidence just by reading the full url of the link. When that is the source given it does come across as an issue, and is the main issue I have with the content; the wording could possibly be solved by drawing from a better source. - Aoidh (talk) 08:46, 3 December 2022 (UTC)
 * Thanks for bringing this up ! As I said earlier, I agree that there should be better sourcing and agree with you entirely that this shouldn't really be what's cited. I'm also not sure that this is the best way to phrase the content in general but my thoughts at the time were mostly surrounding how some people are abused this way and ignoring that might not be what's best. I've reread the thread and mentions that there's already content about the dangers of educational neglect at the end of the lead and it's somewhat vague? It doesn't mention educational neglect explicitly. I don't think indoctrination is the best way to put it. This source could potentially be used for something different ? There's also this . I'm not sure I'm explaining my thoughts the best here, but maybe you kind of get what I'm trying to say? This is more an American-centric context and maybe what I'm thinking about would be more suited to Homeschooling in the United States. Your thoughts?  Clovermoss 🍀  (talk) 11:20, 3 December 2022 (UTC)
 * Unfortunately I can't view the GBooks link it's saying the page is unavailable for viewing (I notice what can be viewed changes so I'll try again later) but I think the Kunzman paper would be a good source, it's been subsequently cited and used as a reference by a few dozen other papers. - Aoidh (talk) 04:56, 4 December 2022 (UTC)
 * @Aoidh The second source is the better of the two anyways. :) Clovermoss 🍀  (talk) 07:24, 4 December 2022 (UTC)
 * With the Kunzman source in mind do you have a proposed wording for the content? The source does seem focused on the homeschooling through a religious lens, I'll try later to find some sources that discuss a more generalized context as well. - Aoidh (talk) 17:20, 4 December 2022 (UTC)

Honestly I'm still trying to decide that myself. Part of the reason I've had difficulty envisoning certain changes to this article is that I don't want to generalize and I'm a bit anxious about where I'd even start in some respects. I'm wondering if maybe I should start improving Homeschooling in the United States and work from there? What are your thoughts on the neutrality tag? I do think statements like "Also known as interest-led or child-led learning, unschooling attempts to follow opportunities as they arise in real life, through which a child will learn without coercion" is more biased towards homeschooling advocacy, if anything. The person who added the tag seemed to think that this article was biased towards criticism. I don't want to go too much the other way, but I do think there could be a better general balance. Do you think I'm on the right track here? I might try to make some bold edits later. Clovermoss 🍀 (talk) 17:43, 4 December 2022 (UTC)
 * So for better or worse I have never had any exposure to homeschooling, good or bad, so I'd like to think I'm approaching the subject from a disinterested perspective. With that in mind the sentence you quoted doesn't seem particularly biased one way or another. I skimmed through a few random passages and the main thing that I saw was that this article learns heavily on using scare quotes. I'm not going to start messing with the article in my current tired state, but I'll at least try to tackle some of that over-reliable of quotation marks in a little while. - Aoidh (talk) 18:18, 4 December 2022 (UTC)
 * I think I was just being picky about the word coercion because it implies students who aren't unschooled are being coerced. Feel free to tackle this whenever you're ready to. I'm interested in doing something eventually myself but I can get indecisive and distracted so we'll see. Clovermoss 🍀  (talk) 18:23, 4 December 2022 (UTC)
 * Ah, I see what you mean. I think changing the word coercion to something like guidance or just removing it outright is fine too. I did come across a sentence which concerned me, namely the sentence in the lede about homeschooled students "have equally or better developed social skills and participate more in cultural and family activities on average than public school students." The source for that statement says "these are subjective questionnaires completed by children’s parents and it is difficult to determine the validity of parents’ views of their own child’s social skills" but the content in the article certainly does not reflect that this is the a survey of how parents viewed their child's socialization skills. I've made an attempt to reword that to reflect the source. It could probably be worded better or even just removed or moved into the article itself instead of the lede, but what it said was certainly not reflective of the actual content of the reference. - Aoidh (talk) 20:10, 4 December 2022 (UTC)
 * I've been busier today than I thought I would be so I didn't really have the time to do anything drastically different, but I did make a few changes. . If there's anything that I did that was controversial, we can discuss it. Clovermoss 🍀  (talk) 04:49, 6 December 2022 (UTC)

Arizona Law Review source
I just found this source: Homeschooling: Parent Rights Absolutism vs. Child Rights to Education & Protection

I haven't read it fully yet, but a quick skim shows that it covers and critiques a lot of the topics covered in this article. It may be useful as a source, and also may point us to other useful sources. Crossroads -talk- 02:40, 22 June 2024 (UTC)

 * Just wanted to pop in and thank you for all your recent edits here! You've done some great work. I meant to do something about the state of the article a year and a half ago and I completely forgot about it. Clovermoss 🍀  (talk) 03:58, 22 June 2024 (UTC)
 * Thank you for the appreciation! I just had to do something about the completely uncritical reiteration of those studies. Based on what I've seen so far, I'm sure a lot remains to be done. Crossroads -talk- 04:06, 22 June 2024 (UTC)
 * That's how I feel about what I did with dynamic pricing today. I'm incredibly busy this weekend but I'll try to remember to take a look at this article again sometime past Tuesday. We can try to collaborate a bit. :) Clovermoss 🍀  (talk) 04:10, 22 June 2024 (UTC)
 * I've tried to improve the history section. I'm not sure how helpful my edits have been but hopefully it's better than it was before. I'm open to respectfully ruthless feedback. Clovermoss 🍀  (talk) 08:03, 25 June 2024 (UTC)

I was also wondering how exactly you wanted to approach collaboration here? I work on a section, you work on a section? Some other method? Clovermoss 🍀 (talk) 21:34, 25 June 2024 (UTC)
 * Thanks for your work on this so far! I think it's very good. I'm sorry to say I honestly don't know how much time I'll have to spend on this article in the near term, but I'll probably keep coming back and tweaking and adding things here and there. I've still watchlisted it, so I'll comment if any disputes crop up, as well as commenting in other talk page discussions. I suggest you make whatever edits strike you as improvements and I will do the same, and they can thus build on each other.
 * I'd suggest we try to get that article-wide neutrality tag off first, if we're satisfied with the lead. Any sections that we still think are unbalanced can be tagged individually. Crossroads -talk- 02:15, 26 June 2024 (UTC)
 * Okay, sounds like a plan. Clovermoss 🍀  (talk) 03:13, 26 June 2024 (UTC)
 * I'd also like to note that my interest in this article may ebb and flow but I don't plan on completely forgetting about it. It's difficult to find the motivation when dealing with an article like this. Anyways, I haven't done much with this source yet but I think it could useful elsewhere. Clovermoss 🍀  (talk) 10:55, 1 July 2024 (UTC)
 * I am much the same way, so I completely understand. Thank you for your latest edits and for the new source to read. Crossroads -talk- 17:21, 2 July 2024 (UTC)

title: Talk:Homeschooling


text:
Humayun

Nasir al-Din Muhammad (6 March 1508 – 27 January 1556), commonly known by his regnal name Humayun, was the second Mughal emperor, who ruled over territory in what is now Eastern Afghanistan, Bangladesh, Northern India, and Pakistan from 1530 to 1540 and again from 1555 to his death in 1556. At the time of his death, the Mughal Empire spanned almost one million square kilometers.

On 26 December 1530, Humayun succeeded his father Babur to the throne of Delhi as ruler of the Mughal territories in the Indian subcontinent. Humayun was an inexperienced ruler when he came to power at the age of 22. His half-brother Kamran Mirza inherited Kabul and Kandahar, the northernmost parts of their father's empire; the two half-brothers became bitter rivals.

Early in his reign, Humayun lost his entire state to Sher Shah Suri but regained it 15 years later with Safavid aid. His return from Persia was accompanied by a large retinue of Persian noblemen, signaling an important change in Mughal court culture. The Central Asian origins of the dynasty were largely overshadowed by the influences of Persian art, architecture, language, and literature. To this day, stone carvings and thousands of Persian manuscripts in India dating from the time of Humayun remain in the subcontinent. Following his return to power, Humayun quickly expanded the Empire, leaving a substantial legacy for his son, Akbar.

Background
Humayun was born as Nasir al-Din Muhammad to Babur's favorite wife Māham Begum on Tuesday 6 March 1508. According to Abul Fazl, Māham was related to the noble family of Sultan Husayn Bayqara, the Timurid ruler of Herat. She was also related to Sheikh Ahmad-e Jami.

The decision of Babur to divide the territories of his empire between two of his sons was unusual in India, although it had been a common Central Asian practice since the time of Genghis Khan. Unlike most monarchies, which practiced primogeniture, the Timurids followed the example of Genghis and did not leave an entire kingdom to the eldest son. Although under that system only a Chingissid could claim sovereignty and Khanal authority, any male Chinggisid within a given sub-branch had an equal right to the throne (though the Timurids were not Chinggisid in their paternal ancestry). While Genghis Khan's Empire had been peacefully divided between his sons upon his death, almost every Chinggisid succession since had resulted in fratricide.

After Timur's death, his territories were divided among Pir Muhammad, Miran Shah, Khalil Sultan and Shah Rukh, which resulted in inter-family warfare. Upon Babur's death, Humayun's territories were the least secure. He had ruled only four years, and not all umarah (nobles) viewed Humayun as the rightful ruler. Indeed, earlier, when Babur had become ill, some of the nobles had tried to install his Brother-in-law, Mahdi Khwaja, as ruler. Although this attempt failed, it was a sign of problems to come.

Early reign
When Humayun came to the throne of the Mughal Empire, several of his brothers revolted against him after he split the empire among them . Another brother Khalil Mirza (1509–1530) supported Humayun but was assassinated. The Emperor commenced construction of a tomb for his brother in 1538, but this was not yet finished when he was forced to flee to Persia. Sher Shah destroyed the structure and no further work was done on it after Humayun's restoration.

Humayun had two major rivals for his lands: Sultan Bahadur of Gujarat to the southwest and Sher Shah Suri (Sher Khan) settled along the river Ganges in Bihar to the east. Humayun's first campaign was to confront Sher Shah Suri. Halfway through this offensive Humayun had to abandon it and concentrate on Gujarat, where a threat from Ahmed Shah had to be met. Humayun was victorious annexing Gujarat, Malwa, Champaner and the great fort of Mandu.

During the first five years of Humayun's reign, Bahadur and Sher Khan extended their rule, although Sultan Bahadur faced pressure in the east from sporadic conflicts with the Portuguese. While the Mughals had obtained firearms via the Ottoman Empire, Bahadur's Gujarat had acquired them through a series of contracts drawn up with the Portuguese, allowing the Portuguese to establish a strategic foothold in northwestern India.

In 1535 Humayun was made aware that the Sultan of Gujarat was planning an assault on the Mughal territories in Bayana with Portuguese aid. Humayun gathered an army and marched on Bahadur. Within a month he had captured the forts of Mandu and Champaner. However, instead of pressing his attack, Humayun ceased the campaign and consolidated his newly conquered territory. Sultan Bahadur, meanwhile escaped and took up refuge with the Portuguese. Like his father, Humayun was a frequent user of opium. In a popular revolt Bahadur Shah recaptured all of Gujarat in 1536 and began an attack on Malwa.

Sher Shah Suri

Shortly after Humayun had marched on Gujarat, Sher Shah Suri saw an opportunity to wrest control of Agra from the Mughals. He began to gather his army together hoping for a rapid and decisive siege of the Mughal capital. Upon hearing this alarming news, Humayun quickly marched his troops back to Agra allowing Bahadur to easily regain control of the territories Humayun had recently taken. In February 1537, however, Bahadur was killed when a botched plan to kidnap the Portuguese viceroy ended in a fire-fight that the Sultan lost. Bahadur's passing caused a power vacuum in Gujarat, which ultimately paved the way for the Mughals to become the region's dominant force.

While Humayun succeeded in protecting Agra from Sher Shah, the second city of the Empire, Gaur the capital of the vilayat of Bengal, was sacked. Humayun's troops had been delayed while trying to take Chunar, a fort occupied by Sher Shah's son, in order to protect his troops from an attack from the rear. The stores of grain at Gauri, the largest in the empire, were emptied, and Humayun arrived to see corpses littering the roads. The vast wealth of Bengal was depleted and brought East, giving Sher Shah a substantial war chest.

Sher Shah withdrew to the east, but Humayun did not follow: instead he "shut himself up for a considerable time in his Harem, and indulged himself in every kind of luxury". Hindal, Humayun's 19-year-old brother, had agreed to aid him in this battle and protect the rear from attack, but he abandoned his position and withdrew to Agra, where he decreed himself acting emperor. When Humayun sent the grand Mufti, Sheikh Buhlul, to reason with him; the Sheikh was killed. Further provoking the rebellion, Hindal ordered that the Khutba, or sermon, in the main mosque be surrounded.

Humayun's other brother, Kamran Mirza, marched from his territories in the Punjab, ostensibly to aid Humayun. However, his return home had treacherous motives as he intended to stake a claim for Humayun's apparently collapsing empire. He brokered a deal with Hindal providing that his brother would cease all acts of disloyalty in return for a share in the new empire, which Kamran would create once Humayun was deposed.

In June 1539 Sher Shah met Humayun in the Battle of Chausa on the banks of the Ganges, near Buxar. This was to become an entrenched battle in which both sides spent a lot of time digging themselves into positions. The major part of the Mughal army, the artillery, was now immobile, and Humayun decided to engage in some diplomacy using Muhammad Aziz as ambassador. Humayun agreed to allow Sher Shah to rule over Bengal and Bihar, but only as provinces granted to him by his Emperor, Humayun, falling short of outright sovereignty. The two rulers also struck a bargain in order to save face: Humayun's troops would charge those of Sher Shah whose forces then retreat in feigned fear. Thus honour would, supposedly, be satisfied.

Once the Army of Humayun had made its charge and Sher Shah's troops made their agreed-upon retreat, the Mughal troops relaxed their defensive preparations and returned to their entrenchments without posting a proper guard. Observing the Mughals' vulnerability, Sher Shah reneged on his earlier agreement. That very night, his army approached the Mughal camp and finding the Mughal troops unprepared with a majority asleep, they advanced and killed most of them. The Emperor survived by swimming across the Ganges using an air-filled "water skin", and quietly returned to Agra. Humayun was assisted across the Ganges by Shams al-Din Muhammad.

In Agra

When Humayun returned to Agra, he found that all three of his brothers were present. Humayun once again not only pardoned his brothers for plotting against him, but even forgave Hindal for his outright betrayal. With his armies travelling at a leisurely pace, Sher Shah was gradually drawing closer and closer to Agra. This was a serious threat to the entire family, but Humayun and Kamran squabbled over how to proceed. Kamran withdrew after Humayun refused to make a quick attack on the approaching enemy, instead opting to build a larger army under his own name.

When Kamran returned to Lahore, Humayun, with his other brothers Askari and Hindal, marched to meet Sher Shah 200 km east of Agra at the battle of Kannauj on 17 May 1540. Humayun was soundly defeated. He retreated to Agra, pursued by Sher Shah, and thence through Delhi to Lahore. Sher Shah's founding of the short-lived Sur Empire, with its capital at Delhi, resulted in Humayun's exile for 15 years in the court of Shah Tahmasp I.

In Lahore
The four brothers were united in Lahore, but every day they were informed that Sher Shah was getting closer and closer. When he reached Sirhind, Humayun sent an ambassador carrying the message "I have left you the whole of Hindustan [i.e. the lands to the East of Punjab, comprising most of the Ganges Valley]. Leave Lahore alone, and let Sirhind be a boundary between you and me." Sher Shah, however, replied "I have left you Kabul. You should go there." Kabul was the capital of the empire of Humayun's brother Kamran, who was far from willing to hand over any of his territories to his brother. Instead, Kamran approached Sher Shah and proposed that he actually revolt against his brother and side with Sher Shah in return for most of the Punjab. Sher Shah dismissed his help, believing it not to be required, though word soon spread to Lahore about the treacherous proposal, and Humayun was urged to make an example of Kamran and kill him. Humayun refused, citing the last words of his father, Babur, "Do nothing against your brothers, even though they may deserve it."

Meeting with the Sikh Guru – Guru Angad Sahib
Humayun visited Guru Angad at around 1540 after Humayun lost the Battle of Kannauj, and thereby the Mughal throne to Sher Shah Suri. According to Sikh hagiographies, when Humayun arrived in Gurdwara Mal Akhara Sahib at Khadur Sahib, Guru Angad was sitting and teaching children. The failure to greet the Emperor immediately angered Humayun. Humayun lashed out but the Guru reminded him that the time when you needed to fight when you lost your throne you ran away and did not fight and now you want to attack a person engaged in prayer. In the Sikh texts written more than a century after the event, Guru Angad is said to have blessed the emperor, and reassured him that someday he will regain the throne.

Withdrawing further
Humayun decided it would be wise to withdraw still further. He and his army rode out through and across the Thar Desert, when the Hindu ruler Rao Maldeo Rathore allied with Sher Shah Suri against the Mughal Empire. In many accounts Humayun mentions how he and his pregnant wife had to trace their steps through the desert at the hottest time of year. Their rations were low, and they had little to eat; even drinking water was a major problem in the desert. When Hamida Bano's horse died, no one would lend the Queen (who was now eight months pregnant) a horse, so Humayun did so himself, resulting in him riding a camel for six kilometres (four miles), although Khaled Beg then offered him his mount. Humayun was later to describe this incident as the lowest point in his life. Humayun asked that his brothers join him as he fell back into Sindh. While the previously rebellious Hindal Mirza remained loyal and was ordered to join his brothers in Kandahar, Kamran Mirza and Askari Mirza instead decided to head to the relative peace of Kabul. This was to be a definitive schism in the family. Humayun headed for Sindh because he expected aid from the Emir of Sindh, Hussein Umrani, whom he had appointed and who owed him his allegiance. Also, his wife Hamida hailed from Sindh; she was the daughter of a prestigious pir family (a pir is an Islamic religious guide) of Persian heritage long settled in Sindh. En route to the Emir's court, Humayun had to break journey because his pregnant wife Hamida was unable to travel further. Humayun sought refuge with the Hindu ruler of the oasis town of Amarkot (now part of Sindh province).

Rana Prasad Rao of Amarkot duly welcomed Humayun into his home and sheltered the refugees for several months. Here, in the household of a Hindu Rajput nobleman, Humayun's wife Hamida Bano, daughter of a Sindhi family, gave birth to the future Emperor Akbar on 15 October 1542. The date of birth is well established because Humayun consulted his astronomer to utilise the astrolabe and check the location of the planets. The infant was the long-awaited heir-apparent to the 34-year-old Humayun and the answer of many prayers. Shortly after the birth, Humayun and his party left Amarkot for Sindh, leaving Akbar behind, who was not ready for the grueling journey ahead in his infancy. He was later adopted by Askari Mirza.

For a change, Humayun was not deceived in the character of the man on whom he has pinned his hopes. Emir Hussein Umrani, ruler of Sindh, welcomed Humayun's presence and was loyal to him, just as he had been loyal to Babur against the renegade Arghuns. While in Sindh, Humayun alongside Hussein Umrani, gathered horses and weapons and formed new alliances that helped regain lost territories. Until finally Humayun had gathered hundreds of Sindhi and Baloch tribesmen alongside his Mughals and then marched towards Kandahar and later Kabul, thousands more gathered by his side as Humayun continually declared himself the rightful Timurid heir of the first Mughal Emperor, Babur.

Retreat to Kabul

After Humayun set out from his expedition in Sindh, along with 300 camels (mostly wild) and 2000 loads of grain, he set off to join his brothers in Kandahar after crossing the Indus River on 11 July 1543 along with the ambition to regain the Mughal Empire and overthrow the Suri dynasty. Among the tribes that had sworn allegiance to Humayun were the Leghari, Magsi, Rind and many others.

In Kamran Mirza's territory, Hindal Mirza had been placed under house arrest in Kabul after refusing to have the Khutba recited in Kamran Mirza's name. His other brother, Askari Mirza, was now ordered to gather an army and march on Humayun. When Humayun received word of the approaching hostile army he decided against facing them, and instead sought refuge elsewhere. Akbar was left behind in camp close to Kandahar, as it was December, too cold and dangerous to include the 14-month-old toddler in the march through the mountains of the Hindu Kush. Askari Mirza took Akbar in, leaving the wives of Kamran and Askari Mirza to raise him. The Akbarnama specifies Kamran Mirza's wife, Sultan Begam.

Once again Humayun turned toward Kandahar where his brother Kamran Mirza was in power, but he received no help and had to seek refuge with the Shah of Persia

Refuge in Persia

Humayun fled to the refuge of the Safavid Empire in Persia, marching with 40 men, his wife Bega Begum, and her companion through mountains and valleys. Among other trials the Imperial party were forced to live on horse meat boiled in the soldiers' helmets. These indignities continued during the month it took them to reach Herat, however after their arrival they were reintroduced to the finer things in life. Upon entering the city his army was greeted with an armed escort, and they were treated to lavish food and clothing. They were given fine accommodations and the roads were cleared and cleaned before them. The Shah, Tahmasp I, unlike Humayun's own family, actually welcomed the Mughal, and treated him as a royal visitor. Here Humayun went sightseeing and was amazed at the Persian artwork and architecture he saw: much of this was the work of the Timurid Sultan Husayn Bayqarah and his ancestor, princess Gauhar Shad, thus he was able to admire the work of his relatives and ancestors at first hand.

The Mughal monarch was introduced to the work of the Persian miniaturists, and Kamaleddin Behzad had two of his pupils join Humayun in his court. Humayun was amazed at their work and asked if they would work for him if he were to regain the sovereignty of Hindustan: they agreed. With so much going on Humayun did not even meet Tahmasp until July, some six months after the former's arrival in Persia. After a lengthy journey from Herat the two met in Qazvin where a large feast and parties were held for the event. The meeting of the two emperors is depicted in a famous wall-painting in the Chehel Sotoun (Forty Columns) palace in Esfahan.

Tahmasp urged that Humayun convert from Sunni to Shia Islam in order to keep himself and several hundred followers alive. Although the Mughals initially disagreed to their conversion they knew that with this outward acceptance of Shi'ism, Tahmasp was eventually prepared to offer Humayun more substantial support. When Humayun's brother, Kamran Mirza, offered to cede Kandahar to the Persians in exchange for Humayun, dead or alive, Tahmasp refused. Instead he staged a celebration, with 300 tents, an imperial Persian carpet, 12 musical bands and "meat of all kinds". Here the Shah announced that all this, and 12,000 elite cavalry were Humayun's to lead an attack on Kamran. All that Tahmasp asked for was that, if Humayun's forces were victorious, Kandahar would be his.

Kandahar and onward

With this Persian Safavid aid Humayun took Kandahar from Askari Mirza after a two-week siege. He noted how the nobles who had served Askari Mirza quickly flocked to serve him, "in very truth the greater part of the inhabitants of the world are like a flock of sheep, wherever one goes the others immediately follow". Kandahar was, as agreed, given to the Shah of Persia who sent his infant son, Murad, as the viceroy. However, the baby soon died and Humayun thought himself strong enough to assume power.

Humayun now prepared to take Kabul, ruled by his brother Kamran Mirza. In the end, there was no actual siege. Kamran Mirza was detested as a leader and as Humayun's Persian army approached the city hundreds of the former's troops changed sides, flocking to join Humayun and swelling his ranks. Kamran Mirza absconded and began building an army outside the city. In November 1545, Hamida and Humayun were reunited with their son Akbar, and held a huge feast. They also held another feast in the child's honour when he was circumcised.

However, while Humayun had a larger army than Kamran Mirza and had the upper hand, on two occasions his poor military judgement allowed the latter to retake Kabul and Kandahar, forcing Humayun to mount further campaigns for their recapture. He might have been aided in this by his reputation for leniency towards the troops who had defended the cities against him, as opposed to Kamran Mirza, whose brief periods of possession were marked by atrocities against the inhabitants who, he supposed, had helped his brother.

His youngest brother, Hindal Mirza, formerly the most disloyal of his siblings, died fighting on his behalf. His brother Askari Mirza was shackled in chains at the behest of his nobles and aides. He was allowed go on Hajj, and died en route in the desert outside Damascus.

Humayun's other brother, Kamran Mirza, had repeatedly sought to have him killed. In 1552 Kamran Mirza attempted to make a pact with Islam Shah, Sher Shah's successor, but was apprehended by a Gakhar. The Gakhars were one of the minority of tribal groups who had consistently remained loyal to their oath to the Mughals. Sultan Adam of the Gakhars handed Kamran Mirza over to Humayun. Humayun, though inclined to forgive Kamran Mirza, was warned that allowing his brother's repeated acts of treachery to go unpunished could foment rebellion amongst his own supporters. So, instead of killing Kamran Mirza, Humayun had him blinded, thereby ending any claim by the latter to the throne. He then sent Kamran Mirza on Hajj, as he hoped to see his brother thereby absolved of his offences. However Kamran Mirza died close to Mecca in the Arabian Peninsula in 1557.

Restoration of the Mughal Empire

In 1535, When Humayun was Governor of Gujarat, he encamped near Cambay (Khambhat). Humayun and his army was robbed and plundered by Kolis of Gujarat.

Sher Shah Suri had died in 1545; his son and successor Islam Shah died in 1554. These two deaths left the dynasty reeling and disintegrating. Three rivals for the throne all marched on Delhi, while in many cities leaders tried to stake a claim for independence. This was a perfect opportunity for the Mughals to march back to India.

The Mughal Emperor Humayun gathered a vast army and attempted the challenging task of retaking the throne in Delhi. Due to the Safavid role in Humayun's army, the vast majority of the army of the Shi'a faith, as one Shaikh Ahmad described to Humayun, "My king, I see the whole of your army are Rafizi...Everywhere the names of your soldiers are of this kind. I find they are all Yar Ali or Kashfi Ali or Haider Ali and I have, not found a single man bearing the names of the other Companions." Humayun placed the army under the leadership of Bairam Khan, a wise move given Humayun's record of military ineptitude, and it turned out to be prescient as Bairam proved himself a great tactician.

Bairam Khan led the army through the Punjab virtually unopposed. The only major battle faced by Humayun's armies was against Sikandar Shah Suri in Sirhind, where Bairam Khan employed a tactic whereby he engaged his enemy in open battle but then retreated quickly in apparent fear. When the enemy followed after them, they were surprised by entrenched defensive positions and were easily annihilated. At the Battle of Sirhind on 22 June 1555, the armies of Sikandar Shah Suri were decisively defeated and the Mughal Empire was reestablished.

After Sirhind, most towns and villages chose to welcome the invading army as it made its way to the capital. On 23 July 1555, Humayun once again sat on Babur's throne in Delhi.

Marriage relations with the Khanzadas
The Gazetteer of Ulwur states: "Soon after Babur's death, his successor, Humayun, was in AD 1540 supplanted by the Pathan Sher Shah, who, in AD 1545, was followed by Islam Shah. During the reign of the latter a battle was fought and lost by the Emperor's troops at Firozpur Jhirka, in Mewat, on which, however, Islam Shah did not loose his hold. Adil Shah, the third of the Pathan interlopers, who succeeded in AD 1552, had to contend for the Empire with the returned Humayun.

In these struggles for the restoration of Babar's dynasty Khanzadas apparently do not figure at all. Humayun seems to have conciliated them by marrying the elder daughter of Jamal Khan, nephew of Babar's opponent, Hasan Khan and, by causing his great minister, Bairam Khan, to marry a younger daughter of the same Mewatti."

Ruling Kashmir

With all of Humayun's brothers now dead, there was no fear of another usurping his throne during his military campaigns. He was also now an established leader and could trust his generals. With this new-found strength Humayun embarked on a series of military campaigns aimed at extending his reign over areas in the east and west of the subcontinent. His sojourn in exile seems to have reduced his reliance, and his military leadership came to imitate the more effective methods that he had observed in Persia.

Character
Edward S. Holden writes; "He was uniformly kind and considerate to his dependents, devotedly attached to his son Akbar, to his friends, and to his turbulent brothers. The misfortunes of his reign arose in great part, from his failure to treat them with rigor." He further writes: "The very defects of his character, which render him less admirable as a successful ruler of nations, make us more fond of him as a man. His renown has suffered in that his reign came between the brilliant conquests of Babur and the beneficent statesmanship of Akbar; but he was not unworthy to be the son of the one and the father of the other."

Stanley Lane-Poole writes in his book Medieval India: "His name meant the winner (Lucky/Conqueror), there is no king in the history to be named as wrong as Humayun", he was of a forgiving nature. He further writes, "He was in fact unfortunate ... Scarcely had he enjoyed his throne for six months in Delhi when he slipped down from the polished steps of his palace and died in his forty-ninth year (Jan. 24, 1556). If there was a possibility of falling, Humayun was not the man to miss it. He tumbled through his life and tumbled out of it."

Humayun ordered the crushing by elephant of an imam he mistakenly believed to be critical of his reign.

Death and legacy

On 24 January 1556, Humayun, with his arms full of books, was descending the staircase from his library Sher Mandal when the muezzin announced the Azaan (the call to prayer). It was his habit, wherever and whenever he heard the summons, to bow his knee in holy reverence. Trying to kneel, he caught his foot in his robe, slipped down several steps and hit his temple on a rugged stone edge. He died three days later. His body was laid to rest in Purana Quila initially, but, because of an attack by Hemu on Delhi and the capture of Purana Qila, Humayun's body was exhumed by the fleeing army and transferred to Kalanaur in Punjab where Akbar was crowned. After young Mughal emperor Akbar defeated and killed Hemu in the Second Battle of Panipat. Humayun's body was buried in Humayun's Tomb in Delhi the first very grand garden tomb in Mughal architecture, setting the precedent later followed by the Taj Mahal and many other Indian monuments. It was commissioned by his favorite and devoted chief wife, Bega Begum.

Akbar later asked his paternal aunt, Gulbadan Begum, to write a biography of his father Humayun, the Humayun nameh (or Humayun-nama), and what she remembered of Babur.

The full title is Ahwal Humayun Padshah Jamah Kardom Gulbadan Begum bint Babur Padshah amma Akbar Padshah. She was only eight when Babur died, and was married at 17, and her work is in simple Persian style.

Unlike other Mughal royal biographies (the Zafarnama of Timur, Baburnama, and his own Akbarnama) no richly illustrated copy has survived, and the work is only known from a single battered and slightly incomplete manuscript, now in the British Library, that emerged in the 1860s. Annette Beveridge published an English translation in 1901, and editions in English and Bengali have been published since 2000.

title: Humayun