Hyperbola

Definitions

As locus of points

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

A hyperbola is a set of points, such that for any point ${\displaystyle P}$ of the set, the absolute difference of the distances ${\displaystyle |P{F}_1|,|P{F}_2|}$ to two fixed points ${\displaystyle F_1,F_2}$ (the foci) is constant, usually denoted by ${\displaystyle 2a,a>0}$: $${\displaystyle H=\left\{P:\left|\left|P{F}_2\right|-\left|P{F}_1\right|\right|=2a\right\}.}$$

The midpoint ${\displaystyle M}$ 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 ${\displaystyle V_1,V_2}$, which have distance ${\displaystyle a}$ to the center. The distance ${\displaystyle c}$ of the foci to the center is called the focal distance or linear eccentricity. The quotient ${\displaystyle \frac{c}{a}}$ is the eccentricity ${\displaystyle e}$.

The equation ${\displaystyle \left|\left|P{F}_2\right|-\left|P{F}_1\right|\right|=2a}$ can be viewed in a different way (see diagram):
If ${\displaystyle c_2}$ is the circle with midpoint ${\displaystyle F_2}$ and radius ${\displaystyle 2a}$, then the distance of a point ${\displaystyle P}$ of the right branch to the circle ${\displaystyle c_2}$ equals the distance to the focus ${\displaystyle F_1}$: $${\displaystyle |P{F}_1|=|P{c}_2|.}$$ ${\displaystyle c_2}$ is called the circular directrix (related to focus ${\displaystyle F_2}$) of the hyperbola. In order to get the left branch of the hyperbola, one has to use the circular directrix related to ${\displaystyle F_1}$. 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

If the xy-coordinate system is rotated about the origin by the angle ${\displaystyle +{45}^{\circ }}$ and new coordinates ${\displaystyle \xi ,\eta }$ are assigned, then ${\displaystyle x=\frac{\xi +\eta }{\sqrt{2}},y=\frac{-\xi +\eta }{\sqrt{2}}}$.
The rectangular hyperbola ${\displaystyle \frac{x^2-y^2}{a^2}=1}$ (whose semi-axes are equal) has the new equation ${\displaystyle \frac{2\xi \eta }{a^2}=1}$. Solving for ${\displaystyle \eta }$ yields ${\displaystyle \eta =\frac{a^2/2}{\xi }.}$

Thus, in an xy-coordinate system the graph of a function ${\displaystyle f:x\mapsto \frac{A}{x},A>0,}$ with equation $${\displaystyle y=\frac{A}{x},A>0,}$$ is a rectangular hyperbola entirely in the first and third quadrants with

• the coordinate axes as asymptotes,
• the line ${\displaystyle y=x}$ as major axis ,
• the center ${\displaystyle (0,0)}$ and the semi-axis ${\displaystyle a=b=\sqrt{2A},}$
• the vertices ${\displaystyle \left(\sqrt{A},\sqrt{A}\right),\left(-\sqrt{A},-\sqrt{A}\right),}$
• the semi-latus rectum and radius of curvature at the vertices ${\displaystyle p=a=\sqrt{2A},}$
• the linear eccentricity ${\displaystyle c=2\sqrt{A}}$ and the eccentricity ${\displaystyle e=\sqrt{2},}$
• the tangent ${\displaystyle y=-\frac{A}{x_0^2}x+2\frac{A}{x_0}}$ at point ${\displaystyle (x_0,A/x_0).}$

A rotation of the original hyperbola by ${\displaystyle -{45}^{\circ }}$ 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 ${\displaystyle +{45}^{\circ }}$ rotation, with equation $${\displaystyle y=-\frac{A}{x},A>0,}$$

• the semi-axes ${\displaystyle a=b=\sqrt{2A},}$
• the line ${\displaystyle y=-x}$ as major axis,
• the vertices ${\displaystyle \left(-\sqrt{A},\sqrt{A}\right),\left(\sqrt{A},-\sqrt{A}\right).}$

Shifting the hyperbola with equation ${\displaystyle y=\frac{A}{x},A\ne 0,}$ so that the new center is ${\displaystyle (c_0,d_0)}$, yields the new equation $${\displaystyle y=\frac{A}{x-c_0}+d_0,}$$ and the new asymptotes are ${\displaystyle x=c_0}$ and ${\displaystyle y=d_0}$. The shape parameters ${\displaystyle a,b,p,c,e}$ 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 ${\displaystyle P}$ of the hyperbola the quotient of the distance to one focus and to the corresponding directrix (see diagram) is equal to the eccentricity: $${\displaystyle \frac{|P{F}_1|}{|P{l}_1|}=\frac{|P{F}_2|}{|P{l}_2|}=e=\frac{c}{a}.}$$ The proof for the pair ${\displaystyle F_1,l_1}$ follows from the fact that ${\displaystyle |P{F}_1{|}^2=(x-c{)}^2+y^2,|P{l}_1{|}^2={\left(x-\frac{a^2}{c}\right)}^2}$ and ${\displaystyle y^2=\frac{b^2}{a^2}{x}^2-b^2}$ satisfy the equation $${\displaystyle |P{F}_1{|}^2-\frac{c^2}{a^2}|P{l}_1{|}^2=0.}$$ 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 ${\displaystyle F}$ (focus), any line ${\displaystyle l}$ (directrix) not through ${\displaystyle F}$ and any real number ${\displaystyle e}$ with ${\displaystyle e>1}$ the set of points (locus of points), for which the quotient of the distances to the point and to the line is ${\displaystyle e}$ $${\displaystyle H=\left\{P|\frac{|PF|}{|Pl|}=e\right\}}$$ is a hyperbola.

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

Proof

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

${\displaystyle (x-f{)}^2+y^2=e^2{\left(x+\frac{f}{e}\right)}^2=(ex+f{)}^2}$ and ${\displaystyle x^2(e^2-1)+2xf(1+e)-y^2=0.}$

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

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

Construction of a directrix

Because of ${\displaystyle c\cdot \frac{a^2}{c}=a^2}$ point ${\displaystyle L_1}$ of directrix ${\displaystyle l_1}$ (see diagram) and focus ${\displaystyle F_1}$ are inverse with respect to the circle inversion at circle ${\displaystyle x^2+y^2=a^2}$ (in diagram green). Hence point ${\displaystyle E_1}$ can be constructed using the theorem of Thales (not shown in the diagram). The directrix ${\displaystyle l_1}$ is the perpendicular to line ${\displaystyle \overline{F_1{F}_2}}$ through point ${\displaystyle E_1}$.

Alternative construction of ${\displaystyle E_1}$: Calculation shows, that point ${\displaystyle E_1}$ is the intersection of the asymptote with its perpendicular through ${\displaystyle F_1}$ (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 ${\displaystyle d_1,d_2}$, which are spheres that touch the cone along circles ${\displaystyle c_1}$, ${\displaystyle c_2}$ and the intersecting (hyperbola) plane at points ${\displaystyle F_1}$ and ${\displaystyle F_2}$. It turns out: ${\displaystyle F_1,F_2}$ are the foci of the hyperbola.

1. Let ${\displaystyle P}$ be an arbitrary point of the intersection curve.
2. The generatrix of the cone containing ${\displaystyle P}$ intersects circle ${\displaystyle c_1}$ at point ${\displaystyle A}$ and circle ${\displaystyle c_2}$ at a point ${\displaystyle B}$.
3. The line segments ${\displaystyle \overline{P{F}_1}}$ and ${\displaystyle \overline{PA}}$ are tangential to the sphere ${\displaystyle d_1}$ and, hence, are of equal length.
4. The line segments ${\displaystyle \overline{P{F}_2}}$ and ${\displaystyle \overline{PB}}$ are tangential to the sphere ${\displaystyle d_2}$ and, hence, are of equal length.
5. The result is: ${\displaystyle |P{F}_1|-|P{F}_2|=|PA|-|PB|=|AB|}$ is independent of the hyperbola point ${\displaystyle P}$, because no matter where point ${\displaystyle P}$ is, ${\displaystyle A,B}$ have to be on circles ${\displaystyle c_1}$, ${\displaystyle c_2}$, and line segment ${\displaystyle AB}$ has to cross the apex. Therefore, as point ${\displaystyle P}$ moves along the red curve (hyperbola), line segment ${\displaystyle \overline{AB}}$ 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 ${\displaystyle F_1,F_2}$ and one of the circular directrices, for example ${\displaystyle c_2}$ (circle with radius ${\displaystyle 2a}$)
2. A ruler is fixed at point ${\displaystyle F_2}$ free to rotate around ${\displaystyle F_2}$. Point ${\displaystyle B}$ is marked at distance ${\displaystyle 2a}$.
3. A string gets its one end pinned at point ${\displaystyle A}$ on the ruler and its length is made ${\displaystyle |AB|}$.
4. The free end of the string is pinned to point ${\displaystyle F_1}$.
5. Take a pen and hold the string tight to the edge of the ruler.
6. Rotating the ruler around ${\displaystyle F_2}$ prompts the pen to draw an arc of the right branch of the hyperbola, because of ${\displaystyle |P{F}_1|=|PB|}$ (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:

Given two pencils ${\displaystyle B(U),B(V)}$ of lines at two points ${\displaystyle U,V}$ (all lines containing ${\displaystyle U}$ and ${\displaystyle V}$, respectively) and a projective but not perspective mapping ${\displaystyle \pi }$ of ${\displaystyle B(U)}$ onto ${\displaystyle B(V)}$, then the intersection points of corresponding lines form a non-degenerate projective conic section.

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

Remarks:

• The subdivision could be extended beyond the points ${\displaystyle A}$ and ${\displaystyle B}$ 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 ${\displaystyle y=\frac{a}{x-b}+c,a\ne 0}$ is uniquely determined by three points ${\displaystyle (x_1,y_1),(x_2,y_2),(x_3,y_3)}$ with different x- and y-coordinates. A simple way to determine the shape parameters ${\displaystyle a,b,c}$ uses the inscribed angle theorem for hyperbolas:

In order to measure an angle between two lines with equations ${\displaystyle y=m_{1}x+d_1,y=m_{2}x+d_2,m_1,m_2\ne 0}$ in this context one uses the quotient $${\displaystyle \frac{m_1}{m_2}.}$$

Analogous to the inscribed angle theorem for circles one gets the

Inscribed angle theorem for hyperbolas—For four points ${\displaystyle P_i=(x_i,y_i),i=1,2,3,4,x_i\ne x_k,y_i\ne y_k,i\ne k}$ (see diagram) the following statement is true:

The four points are on a hyperbola with equation ${\displaystyle y=\frac{a}{x-b}+c}$ if and only if the angles at ${\displaystyle P_3}$ and ${\displaystyle P_4}$ are equal in the sense of the measurement above. That means if $${\displaystyle \frac{(y_4-y_1)}{(x_4-x_1)}\frac{(x_4-x_2)}{(y_4-y_2)}=\frac{(y_3-y_1)}{(x_3-x_1)}\frac{(x_3-x_2)}{(y_3-y_2)}}$$

The proof can be derived by straightforward calculation. If the points are on a hyperbola, one can assume the hyperbola's equation is ${\displaystyle y=a/x}$.

A consequence of the inscribed angle theorem for hyperbolas is the

3-point-form of a hyperbola's equation—The equation of the hyperbola determined by 3 points ${\displaystyle P_i=(x_i,y_i),i=1,2,3,x_i\ne x_k,y_i\ne y_k,i\ne k}$ is the solution of the equation $${\displaystyle \frac{(y-y_1)}{(x-x_1)}\frac{(x-x_2)}{(y-y_2)}=\frac{(y_3-y_1)}{(x_3-x_1)}\frac{(x_3-x_2)}{(y_3-y_2)}}$$ for ${\displaystyle y}$.

As an affine image of the unit hyperbola x2 − y2 = 1

Another definition of a hyperbola uses affine transformations:

Any hyperbola is the affine image of the unit hyperbola with equation ${\displaystyle x^2-y^2=1}$.

Parametric representation

An affine transformation of the Euclidean plane has the form ${\displaystyle \overrightarrow{x}\to {\overrightarrow{f}}_0+A\overrightarrow{x}}$, where ${\displaystyle A}$ is a regular matrix (its determinant is not 0) and ${\displaystyle {\overrightarrow{f}}_0}$ is an arbitrary vector. If ${\displaystyle {\overrightarrow{f}}_1,{\overrightarrow{f}}_2}$ are the column vectors of the matrix ${\displaystyle A}$, the unit hyperbola ${\displaystyle (\pm \cosh (t),\sinh (t)),t\in \mathbb{R},}$ is mapped onto the hyperbola

$${\displaystyle \overrightarrow{x}=\overrightarrow{p}(t)={\overrightarrow{f}}_0\pm {\overrightarrow{f}}_1\cosh t+{\overrightarrow{f}}_2\sinh t.}$$

${\displaystyle {\overrightarrow{f}}_0}$ is the center, ${\displaystyle {\overrightarrow{f}}_0+{\overrightarrow{f}}_1}$ a point of the hyperbola and ${\displaystyle {\overrightarrow{f}}_2}$ a tangent vector at this point.

Vertices

In general the vectors ${\displaystyle {\overrightarrow{f}}_1,{\overrightarrow{f}}_2}$ are not perpendicular. That means, in general ${\displaystyle {\overrightarrow{f}}_0\pm {\overrightarrow{f}}_1}$ are not the vertices of the hyperbola. But ${\displaystyle {\overrightarrow{f}}_1\pm {\overrightarrow{f}}_2}$ point into the directions of the asymptotes. The tangent vector at point ${\displaystyle \overrightarrow{p}(t)}$ is $${\displaystyle {\overrightarrow{p}}^{\prime }(t)={\overrightarrow{f}}_1\sinh t+{\overrightarrow{f}}_2\cosh t.}$$ Because at a vertex the tangent is perpendicular to the major axis of the hyperbola one gets the parameter ${\displaystyle t_0}$ of a vertex from the equation $${\displaystyle {\overrightarrow{p}}^{\prime }(t)\cdot \left(\overrightarrow{p}(t)-{\overrightarrow{f}}_0\right)=\left({\overrightarrow{f}}_1\sinh t+{\overrightarrow{f}}_2\cosh t\right)\cdot \left({\overrightarrow{f}}_1\cosh t+{\overrightarrow{f}}_2\sinh t\right)=0}$$ and hence from $${\displaystyle \coth (2t_0)=-\frac{{\overrightarrow{f}}_1^2+{\overrightarrow{f}}_2^2}{2{\overrightarrow{f}}_1\cdot {\overrightarrow{f}}_2},}$$ which yields

$${\displaystyle t_0=\frac{1}{4}\ln \frac{{\left({\overrightarrow{f}}_1-{\overrightarrow{f}}_2\right)}^2}{{\left({\overrightarrow{f}}_1+{\overrightarrow{f}}_2\right)}^2}.}$$

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

The two vertices of the hyperbola are ${\displaystyle {\overrightarrow{f}}_0\pm \left({\overrightarrow{f}}_1\cosh t_0+{\overrightarrow{f}}_2\sinh t_0\right).}$

Implicit representation

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

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 ${\displaystyle \overrightarrow{f}{}_0,\overrightarrow{f}{}_1,\overrightarrow{f}{}_2}$ to be vectors in space.

As an affine image of the hyperbola y = 1/x

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

$${\displaystyle \overrightarrow{x}=\overrightarrow{p}(t)={\overrightarrow{f}}_0+{\overrightarrow{f}}_{1}t+{\overrightarrow{f}}_2\frac{1}{t},t\ne 0.}$$

${\displaystyle M:{\overrightarrow{f}}_0}$ is the center of the hyperbola, the vectors ${\displaystyle {\overrightarrow{f}}_1,{\overrightarrow{f}}_2}$ have the directions of the asymptotes and ${\displaystyle {\overrightarrow{f}}_1+{\overrightarrow{f}}_2}$ is a point of the hyperbola. The tangent vector is $${\displaystyle {\overrightarrow{p}}^{\prime }(t)={\overrightarrow{f}}_1-{\overrightarrow{f}}_2\frac{1}{t^2}.}$$ At a vertex the tangent is perpendicular to the major axis. Hence $${\displaystyle {\overrightarrow{p}}^{\prime }(t)\cdot \left(\overrightarrow{p}(t)-{\overrightarrow{f}}_0\right)=\left({\overrightarrow{f}}_1-{\overrightarrow{f}}_2\frac{1}{t^2}\right)\cdot \left({\overrightarrow{f}}_{1}t+{\overrightarrow{f}}_2\frac{1}{t}\right)={\overrightarrow{f}}_1^{2}t-{\overrightarrow{f}}_2^2\frac{1}{t^3}=0}$$ and the parameter of a vertex is

$${\displaystyle t_0=\pm \sqrt[4]{\frac{{\overrightarrow{f}}_2^2}{{\overrightarrow{f}}_1^2}}.}$$

${\displaystyle \left|\overrightarrow{f}{}_1\right|=\left|\overrightarrow{f}{}_2\right|}$ is equivalent to ${\displaystyle t_0=\pm 1}$ and ${\displaystyle {\overrightarrow{f}}_0\pm ({\overrightarrow{f}}_1+{\overrightarrow{f}}_2)}$ 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: $${\displaystyle {\overrightarrow{p}}^{\prime }(t)=\frac{1}{t}\left({\overrightarrow{f}}_{1}t-{\overrightarrow{f}}_2\frac{1}{t}\right).}$$ This means that

the diagonal ${\displaystyle AB}$ of the parallelogram ${\displaystyle M:{\overrightarrow{f}}_0,A={\overrightarrow{f}}_0+{\overrightarrow{f}}_{1}t,B:{\overrightarrow{f}}_0+{\overrightarrow{f}}_2\frac{1}{t},P:{\overrightarrow{f}}_0+{\overrightarrow{f}}_{1}t+{\overrightarrow{f}}_2\frac{1}{t}}$ is parallel to the tangent at the hyperbola point ${\displaystyle P}$ (see diagram).

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.

Area of the grey parallelogram

The area of the grey parallelogram ${\displaystyle MAPB}$ in the above diagram is $${\displaystyle \text{Area}=\left|det\left(t{\overrightarrow{f}}_1,\frac{1}{t}{\overrightarrow{f}}_2\right)\right|=\left|det\left({\overrightarrow{f}}_1,{\overrightarrow{f}}_2\right)\right|=\cdots =\frac{ab}{2}}$$ and hence independent of point ${\displaystyle P}$. The last equation follows from a calculation for the case, where ${\displaystyle P}$ is a vertex and the hyperbola in its canonical form ${\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1.}$

Point construction

For a hyperbola with parametric representation ${\displaystyle \overrightarrow{x}=\overrightarrow{p}(t)={\overrightarrow{f}}_{1}t+{\overrightarrow{f}}_2\frac{1}{t}}$ (for simplicity the center is the origin) the following is true:

For any two points ${\displaystyle P_1:{\overrightarrow{f}}_1{t}_1+{\overrightarrow{f}}_2\frac{1}{t_1},P_2:{\overrightarrow{f}}_1{t}_2+{\overrightarrow{f}}_2\frac{1}{t_2}}$ the points

$${\displaystyle A:\overrightarrow{a}={\overrightarrow{f}}_1{t}_1+{\overrightarrow{f}}_2\frac{1}{t_2},B:\overrightarrow{b}={\overrightarrow{f}}_1{t}_2+{\overrightarrow{f}}_2\frac{1}{t_1}}$$

are collinear with the center of the hyperbola (see diagram).

The simple proof is a consequence of the equation ${\displaystyle \frac{1}{t_1}\overrightarrow{a}=\frac{1}{t_2}\overrightarrow{b}}$.

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 ${\displaystyle {\overrightarrow{f}}_1,{\overrightarrow{f}}_2}$ 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 ${\displaystyle \pm ({\overrightarrow{f}}_1+{\overrightarrow{f}}_2)}$ are the vertices, ${\displaystyle \pm ({\overrightarrow{f}}_1-{\overrightarrow{f}}_2)}$ span the minor axis and one gets ${\displaystyle |{\overrightarrow{f}}_1+{\overrightarrow{f}}_2|=a}$ and ${\displaystyle |{\overrightarrow{f}}_1-{\overrightarrow{f}}_2|=b}$.

For the intersection points of the tangent at point ${\displaystyle \overrightarrow{p}(t_0)={\overrightarrow{f}}_1{t}_0+{\overrightarrow{f}}_2\frac{1}{t_0}}$ with the asymptotes one gets the points $${\displaystyle C=2t_0{\overrightarrow{f}}_1,D=\frac{2}{t_0}{\overrightarrow{f}}_2.}$$ The area of the triangle ${\displaystyle M,C,D}$ can be calculated by a 2 × 2 determinant: $${\displaystyle A=\frac{1}{2}|det\left(2t_0{\overrightarrow{f}}_1,\frac{2}{t_0}{\overrightarrow{f}}_2\right)|=2|det\left({\overrightarrow{f}}_1,{\overrightarrow{f}}_2\right)|}$$ (see rules for determinants). ${\displaystyle \left|det({\overrightarrow{f}}_1,{\overrightarrow{f}}_2)\right|}$ is the area of the rhombus generated by ${\displaystyle {\overrightarrow{f}}_1,{\overrightarrow{f}}_2}$. The area of a rhombus is equal to one half of the product of its diagonals. The diagonals are the semi-axes ${\displaystyle a,b}$ of the hyperbola. Hence:

The area of the triangle ${\displaystyle MCD}$ is independent of the point of the hyperbola: ${\displaystyle A=ab.}$

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

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

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 ${\displaystyle (x,y)}$ in the plane,

$${\displaystyle A_{xx}{x}^2+2A_{xy}xy+A_{yy}{y}^2+2B_{x}x+2B_{y}y+C=0,}$$

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

$${\displaystyle D:=\left|\begin{array}{cc}{A}_{xx}& A_{xy}\\ A_{xy}& A_{yy}\end{array}\right|<0.}$$

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:

$${\displaystyle \mathrm{\Delta }:=\left|\begin{array}{ccc}{A}_{xx}& A_{xy}& B_x\\ A_{xy}& A_{yy}& B_y\\ B_x& B_y& C\end{array}\right|=0.}$$

This determinant ${\displaystyle \mathrm{\Delta }}$ is sometimes called the discriminant of the conic section.

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

$${\displaystyle \begin{array}{cccc}{A}_{xx}& =-a^2{\sin }^2\theta +b^2{\cos }^2\theta ,& B_x& =-A_{xx}{x}_{\circ }-A_{xy}{y}_{\circ },\\ A_{yy}& =-a^2{\cos }^2\theta +b^2{\sin }^2\theta ,& B_y& =-A_{xy}{x}_{\circ }-A_{yy}{y}_{\circ },\\ A_{xy}& =\left(a^2+b^2\right)\sin \theta \cos \theta ,& C& =A_{xx}{x}_{\circ }^2+2A_{xy}{x}_{\circ }{y}_{\circ }+A_{yy}{y}_{\circ }^2-a^2{b}^2.\end{array}}$$

These expressions can be derived from the canonical equation

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

by a translation and rotation of the coordinates ${\displaystyle (x,y)}$:

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

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

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

$${\displaystyle \begin{array}{cc}{x}_c& =-\frac{1}{D}\left|\begin{array}{cc}{B}_x& A_{xy}\\ B_y& A_{yy}\end{array}\right|,\\ y_c& =-\frac{1}{D}\left|\begin{array}{cc}{A}_{xx}& B_x\\ A_{xy}& B_y\end{array}\right|.\end{array}}$$

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

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

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

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

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

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

The major and minor semiaxes ${\displaystyle a}$ and ${\displaystyle b}$ are defined by the equations $${\displaystyle \begin{array}{cc}{a}^2& =-\frac{\mathrm{\Delta }}{{\lambda }_{1}D}=-\frac{\mathrm{\Delta }}{{\lambda }_1^2{\lambda }_2},\\ b^2& =-\frac{\mathrm{\Delta }}{{\lambda }_{2}D}=-\frac{\mathrm{\Delta }}{{\lambda }_1{\lambda }_2^2},\end{array}}$$

where ${\displaystyle {\lambda }_1}$ and ${\displaystyle {\lambda }_2}$ are the roots of the quadratic equation

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

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

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

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

$${\displaystyle Ex+Fy+G=0}$$

where ${\displaystyle E,}$ ${\displaystyle F,}$ and ${\displaystyle G}$ are defined by

$${\displaystyle \begin{array}{cc}E& =A_{xx}{x}_0+A_{xy}{y}_0+B_x,\\ F& =A_{xy}{x}_0+A_{yy}{y}_0+B_y,\\ G& =B_x{x}_0+B_y{y}_0+C.\end{array}}$$

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

$${\displaystyle F(x-x_0)-E(y-y_0)=0.}$$

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

From the equation

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

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

$${\displaystyle r_1-r_2=2a,}$$

and for a point on the left branch,

$${\displaystyle r_2-r_1=2a.}$$

This can be proved as follows:

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

$${\displaystyle r_1^2=(x+ae{)}^2+y^2=x^2+2xae+a^2{e}^2+\left(x^2-a^2\right)\left(e^2-1\right)=(ex+a{)}^2.}$$

To the right focal point the distance is

$${\displaystyle r_2^2=(x-ae{)}^2+y^2=x^2-2xae+a^2{e}^2+\left(x^2-a^2\right)\left(e^2-1\right)=(ex-a{)}^2.}$$

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

$${\displaystyle \begin{array}{cc}{r}_1& =ex+a,\\ r_2& =ex-a.\end{array}}$$

Subtracting these equations one gets

$${\displaystyle r_1-r_2=2a.}$$

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

$${\displaystyle \begin{array}{cc}{r}_1& =-ex-a,\\ r_2& =-ex+a.\end{array}}$$

Subtracting these equations one gets

$${\displaystyle r_2-r_1=2a.}$$

In Cartesian coordinates

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 ${\displaystyle F_1=(c,0),F_2=(-c,0)}$,

the vertices are ${\displaystyle V_1=(a,0),V_2=(-a,0)}$.

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

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

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: ${\displaystyle (a,0),(-a,0)}$. The two points ${\displaystyle (0,b),(0,-b)}$ 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

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

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

With the help of the second figure one can see that

${\displaystyle (1)}$ The perpendicular distance from a focus to either asymptote is ${\displaystyle b}$ (the semi-minor axis).

From the Hesse normal form ${\displaystyle \frac{bx\pm ay}{\sqrt{a^2+b^2}}=0}$ of the asymptotes and the equation of the hyperbola one gets:

${\displaystyle (2)}$ The product of the distances from a point on the hyperbola to both the asymptotes is the constant ${\displaystyle \frac{a^2{b}^2}{a^2+b^2},}$ which can also be written in terms of the eccentricity e as ${\displaystyle {\left(\frac{b}{e}\right)}^2.}$

From the equation ${\displaystyle y=\pm \frac{b}{a}\sqrt{x^2-a^2}}$ of the hyperbola (above) one can derive:

${\displaystyle (3)}$ The product of the slopes of lines from a point P to the two vertices is the constant ${\displaystyle b^2/a^2.}$

In addition, from (2) above it can be shown that

${\displaystyle (4)}$ The product of the distances from a point on the hyperbola to the asymptotes along lines parallel to the asymptotes is the constant ${\displaystyle \frac{a^2+b^2}{4}.}$

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 ${\displaystyle p}$. A calculation shows $${\displaystyle p=\frac{b^2}{a}.}$$ The semi-latus rectum ${\displaystyle p}$ 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 ${\displaystyle (x_0,y_0)}$ is to implicitly differentiate the equation ${\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1}$ of the hyperbola. Denoting dy/dx as y′, this produces $${\displaystyle \frac{2x}{a^2}-\frac{2y{y}^{\prime }}{b^2}=0\Rightarrow y^{\prime }=\frac{x}{y}\frac{b^2}{a^2}\Rightarrow y=\frac{x_0}{y_0}\frac{b^2}{a^2}(x-x_0)+y_0.}$$ With respect to ${\displaystyle \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=1}$, the equation of the tangent at point ${\displaystyle (x_0,y_0)}$ is $${\displaystyle \frac{x_0}{a^2}x-\frac{y_0}{b^2}y=1.}$$

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

Parametric representation with hyperbolic sine/cosine

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

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

Conjugate hyperbola

For the hyperbola ${\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1}$, change the sign on the right to obtain the equation of the conjugate hyperbola:

${\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=-1}$ or equivalently:${\displaystyle \frac{y^2}{b^2}-\frac{x^2}{a^2}=1}$

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.

${\displaystyle xy=c^2}$ and ${\displaystyle xy=-c^2}$ also specify conjugate hyperbolas.

In polar coordinates

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 toward the origin of the "canonical coordinate system" as illustrated in the first diagram.

In this case the angle ${\displaystyle \phi }$ is called true anomaly.

Relative to this coordinate system one has that

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

and

$${\displaystyle -\arccos \left(-\frac{1}{e}\right)<\phi <\arccos \left(-\frac{1}{e}\right).}$$

Origin at the center

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

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

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

Eccentricity

When using polar coordinates, the eccentricity of the hyperbola can be expressed as ${\displaystyle \sec {\phi }_{\text{max}}}$ where ${\displaystyle {\phi }_{\text{max}}}$ is the limit of the angular coordinate. As ${\displaystyle \phi }$ approaches this limit, r approaches infinity and the denominator in either of the equations noted above approaches zero, hence:

$${\displaystyle e^2{\cos }^2{\phi }_{\text{max}}-1=0}$$

$${\displaystyle 1\pm e\cos {\phi }_{\text{max}}=0}$$

$${\displaystyle ⟹e=\sec {\phi }_{\text{max}}}$$

Parametric equations

A hyperbola with equation ${\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1}$ can be described by several parametric equations:

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