Hyperbola

Hyperbola: definition by the distances of points to two fixed points (foci)

Hyperbola: definition with circular directrix

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 ${\displaystyle |PF_{1}|,\,|PF_{2}|}$ to two fixed points ${\displaystyle F_{1},F_{2}}$ (the foci) is constant, usually denoted by ${\displaystyle 2a,\,a>0}$:[6]

$${\displaystyle H=\left\{P:\left|\left|PF_{2}\right|-\left|PF_{1}\right|\right|=2a\right\}.}$$

The midpoint ${\displaystyle M}$ of the line segment joining the foci is called the center of the hyperbola.[7] 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 ${\displaystyle {\tfrac {c}{a}}}$ is the eccentricity ${\displaystyle e}$.

The equation ${\displaystyle \left|\left|PF_{2}\right|-\left|PF_{1}\right|\right|=2a}$ 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 ${\displaystyle F_{1}}$:

$${\displaystyle |PF_{1}|=|Pc_{2}|.}$$

${\displaystyle c_{2}}$ is called the circular directrix (related to focus ${\displaystyle F_{2}}$) of the hyperbola.[8][9] 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.

Rotating the coordinate system in order to describe a rectangular hyperbola as graph of a function

Three rectangular hyperbolas ${\displaystyle y=A/x}$ 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 ${\displaystyle +45^{\circ }}$ and new coordinates ${\displaystyle \xi ,\eta }$ are assigned, then ${\displaystyle x={\tfrac {\xi +\eta }{\sqrt {2}}},\;y={\tfrac {-\xi +\eta }{\sqrt {2}}}}$.
The rectangular hyperbola ${\displaystyle {\tfrac {x^{2}-y^{2}}{a^{2}}}=1}$ (whose semi-axes are equal) has the new equation ${\displaystyle {\tfrac {2\xi \eta }{a^{2}}}=1}$. Solving for ${\displaystyle \eta }$ yields ${\displaystyle \eta ={\tfrac {a^{2}/2}{\xi }}\ .}$

Thus, in an xy-coordinate system the graph of a function ${\displaystyle f:x\mapsto {\tfrac {A}{x}},\;A>0\;,}$ 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 ${\displaystyle y=-{\tfrac {A}{x_{0}^{2}}}x+2{\tfrac {A}{x_{0}}}}$ 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 ${\displaystyle y={\frac {A}{x}},\ A\neq 0\ ,}$ 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.

Hyperbola: directrix property

Hyperbola: definition with directrix property

The two lines at distance ${\textstyle d={\frac {a^{2}}{c}}}$ 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:

$${\displaystyle {\frac {|PF_{1}|}{|Pl_{1}|}}={\frac {|PF_{2}|}{|Pl_{2}|}}=e={\frac {c}{a}}\,.}$$

The proof for the pair ${\displaystyle F_{1},l_{1}}$ follows from the fact that ${\displaystyle |PF_{1}|^{2}=(x-c)^{2}+y^{2},\ |Pl_{1}|^{2}=\left(x-{\tfrac {a^{2}}{c}}\right)^{2}}$ and ${\displaystyle y^{2}={\tfrac {b^{2}}{a^{2}}}x^{2}-b^{2}}$ satisfy the equation

$${\displaystyle |PF_{1}|^{2}-{\frac {c^{2}}{a^{2}}}|Pl_{1}|^{2}=0\ .}$$

The second case is proven analogously.

Pencil of conics with a common vertex and common semi latus rectum

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 ${\displaystyle e}$

$${\displaystyle H=\left\{P\,{\Biggr |}\,{\frac {|PF|}{|Pl|}}=e\right\}}$$

is a hyperbola.

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

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

${\displaystyle (x-f)^{2}+y^{2}=e^{2}\left(x+{\tfrac {f}{e}}\right)^{2}=(ex+f)^{2}}$ 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 ${\displaystyle e^{2}-1={\tfrac {b^{2}}{a^{2}}},{\text{ and }}\ p={\tfrac {b^{2}}{a}}}$, 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 ${\displaystyle a,b}$.

Hyperbola: construction of a directrix

Because of ${\displaystyle c\cdot {\tfrac {a^{2}}{c}}=a^{2}}$ 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).

Hyperbola (red): two views of a cone and two Dandelin spheres d₁, d₂

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 ${\displaystyle {\overline {PF_{1}}}}$ and ${\displaystyle {\overline {PA}}}$ are tangential to the sphere ${\displaystyle d_{1}}$ and, hence, are of equal length.
4. The line segments ${\displaystyle {\overline {PF_{2}}}}$ and ${\displaystyle {\overline {PB}}}$ are tangential to the sphere ${\displaystyle d_{2}}$ and, hence, are of equal length.
5. The result is: ${\displaystyle |PF_{1}|-|PF_{2}|=|PA|-|PB|=|AB|}$ 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.

Hyperbola: 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:[10]

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

Hyperbola: Steiner generation

Hyperbola y = 1/x: Steiner generation

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

Hyperbola: inscribed angle theorem

A hyperbola with equation ${\displaystyle y={\tfrac {a}{x-b}}+c,\ a\neq 0}$ 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 ${\displaystyle y=m_{1}x+d_{1},\ y=m_{2}x+d_{2}\ ,m_{1},m_{2}\neq 0}$ 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[11][12] — For four points ${\displaystyle P_{i}=(x_{i},y_{i}),\ i=1,2,3,4,\ x_{i}\neq x_{k},y_{i}\neq y_{k},i\neq k}$ (see diagram) the following statement is true:

The four points are on a hyperbola with equation ${\displaystyle y={\tfrac {a}{x-b}}+c}$ 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 ${\displaystyle P_{i}=(x_{i},y_{i}),\ i=1,2,3,\ x_{i}\neq x_{k},y_{i}\neq y_{k},i\neq k}$ is the solution of the equation

$${\displaystyle {\frac {({\color {red}y}-y_{1})}{({\color {green}x}-x_{1})}}{\frac {({\color {green}x}-x_{2})}{({\color {red}y}-y_{2})}}={\frac {(y_{3}-y_{1})}{(x_{3}-x_{1})}}{\frac {(x_{3}-x_{2})}{(y_{3}-y_{2})}}}$$

for ${\displaystyle {\color {red}y}}$.

Hyperbola as an affine image of the unit hyperbola

Another definition of a hyperbola uses affine transformations:

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

An affine transformation of the Euclidean plane has the form ${\displaystyle {\vec {x}}\to {\vec {f}}_{0}+A{\vec {x}}}$, where ${\displaystyle A}$ is a regular matrix (its determinant is not 0) and ${\displaystyle {\vec {f}}_{0}}$ is an arbitrary vector. If ${\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}}$ 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

$${\displaystyle {\vec {x}}={\vec {p}}(t)={\vec {f}}_{0}\pm {\vec {f}}_{1}\cosh t+{\vec {f}}_{2}\sinh t\ .}$$

${\displaystyle {\vec {f}}_{0}}$ is the center, ${\displaystyle {\vec {f}}_{0}+{\vec {f}}_{1}}$ a point of the hyperbola and ${\displaystyle {\vec {f}}_{2}}$ a tangent vector at this point.

In general the vectors ${\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}}$ are not perpendicular. That means, in general ${\displaystyle {\vec {f}}_{0}\pm {\vec {f}}_{1}}$ are not the vertices of the hyperbola. But ${\displaystyle {\vec {f}}_{1}\pm {\vec {f}}_{2}}$ point into the directions of the asymptotes. The tangent vector at point ${\displaystyle {\vec {p}}(t)}$ is

$${\displaystyle {\vec {p}}'(t)={\vec {f}}_{1}\sinh t+{\vec {f}}_{2}\cosh t\ .}$$

Because at a vertex the tangent is perpendicular to the major axis of the hyperbola one gets the parameter ${\displaystyle t_{0}}$ of a vertex from the equation

$${\displaystyle {\vec {p}}'(t)\cdot \left({\vec {p}}(t)-{\vec {f}}_{0}\right)=\left({\vec {f}}_{1}\sinh t+{\vec {f}}_{2}\cosh t\right)\cdot \left({\vec {f}}_{1}\cosh t+{\vec {f}}_{2}\sinh t\right)=0}$$

and hence from

$${\displaystyle \coth(2t_{0})=-{\tfrac {{\vec {f}}_{1}^{\,2}+{\vec {f}}_{2}^{\,2}}{2{\vec {f}}_{1}\cdot {\vec {f}}_{2}}}\ ,}$$

which yields

$${\displaystyle t_{0}={\tfrac {1}{4}}\ln {\tfrac {\left({\vec {f}}_{1}-{\vec {f}}_{2}\right)^{2}}{\left({\vec {f}}_{1}+{\vec {f}}_{2}\right)^{2}}}.}$$

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

The two vertices of the hyperbola are ${\displaystyle {\vec {f}}_{0}\pm \left({\vec {f}}_{1}\cosh t_{0}+{\vec {f}}_{2}\sinh t_{0}\right).}$

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

$${\displaystyle \det \left({\vec {x}}\!-\!{\vec {f}}\!_{0},{\vec {f}}\!_{2}\right)^{2}-\det \left({\vec {f}}\!_{1},{\vec {x}}\!-\!{\vec {f}}\!_{0}\right)^{2}-\det \left({\vec {f}}\!_{1},{\vec {f}}\!_{2}\right)^{2}=0.}$$

The definition of a hyperbola in this section gives a parametric representation of an arbitrary hyperbola, even in space, if one allows ${\displaystyle {\vec {f}}\!_{0},{\vec {f}}\!_{1},{\vec {f}}\!_{2}}$ to be vectors in space.

Hyperbola as affine image of 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 ${\displaystyle y=1/x\,}$:

$${\displaystyle {\vec {x}}={\vec {p}}(t)={\vec {f}}_{0}+{\vec {f}}_{1}t+{\vec {f}}_{2}{\tfrac {1}{t}},\quad t\neq 0\,.}$$

${\displaystyle M:{\vec {f}}_{0}}$ is the center of the hyperbola, the vectors ${\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}}$ have the directions of the asymptotes and ${\displaystyle {\vec {f}}_{1}+{\vec {f}}_{2}}$ is a point of the hyperbola. The tangent vector is

$${\displaystyle {\vec {p}}'(t)={\vec {f}}_{1}-{\vec {f}}_{2}{\tfrac {1}{t^{2}}}.}$$

At a vertex the tangent is perpendicular to the major axis. Hence

$${\displaystyle {\vec {p}}'(t)\cdot \left({\vec {p}}(t)-{\vec {f}}_{0}\right)=\left({\vec {f}}_{1}-{\vec {f}}_{2}{\tfrac {1}{t^{2}}}\right)\cdot \left({\vec {f}}_{1}t+{\vec {f}}_{2}{\tfrac {1}{t}}\right)={\vec {f}}_{1}^{2}t-{\vec {f}}_{2}^{2}{\tfrac {1}{t^{3}}}=0}$$

and the parameter of a vertex is

$${\displaystyle t_{0}=\pm {\sqrt[{4}]{\frac {{\vec {f}}_{2}^{2}}{{\vec {f}}_{1}^{2}}}}.}$$

${\displaystyle \left|{\vec {f}}\!_{1}\right|=\left|{\vec {f}}\!_{2}\right|}$ is equivalent to ${\displaystyle t_{0}=\pm 1}$ and ${\displaystyle {\vec {f}}_{0}\pm ({\vec {f}}_{1}+{\vec {f}}_{2})}$ 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: asymptotes and P given → tangent

The tangent vector can be rewritten by factorization:

$${\displaystyle {\vec {p}}'(t)={\tfrac {1}{t}}\left({\vec {f}}_{1}t-{\vec {f}}_{2}{\tfrac {1}{t}}\right)\ .}$$

This means that

the diagonal ${\displaystyle AB}$ of the parallelogram ${\displaystyle M:\ {\vec {f}}_{0},\ A={\vec {f}}_{0}+{\vec {f}}_{1}t,\ B:\ {\vec {f}}_{0}+{\vec {f}}_{2}{\tfrac {1}{t}},\ P:\ {\vec {f}}_{0}+{\vec {f}}_{1}t+{\vec {f}}_{2}{\tfrac {1}{t}}}$ 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.[13]

Area of the grey parallelogram

The area of the grey parallelogram ${\displaystyle MAPB}$ in the above diagram is

$${\displaystyle {\text{Area}}=\left|\det \left(t{\vec {f}}_{1},{\tfrac {1}{t}}{\vec {f}}_{2}\right)\right|=\left|\det \left({\vec {f}}_{1},{\vec {f}}_{2}\right)\right|=\cdots ={\frac {a^{2}+b^{2}}{4}}}$$

and hence independent of point ${\displaystyle P}$. The last equation follows from a calculation for the case, where ${\displaystyle P}$ is a vertex and the hyperbola in its canonical form ${\displaystyle {\tfrac {x^{2}}{a^{2}}}-{\tfrac {y^{2}}{b^{2}}}=1\,.}$

Point construction: asymptotes and P₁ are given → P₂

For a hyperbola with parametric representation ${\displaystyle {\vec {x}}={\vec {p}}(t)={\vec {f}}_{1}t+{\vec {f}}_{2}{\tfrac {1}{t}}}$ (for simplicity the center is the origin) the following is true:

For any two points ${\displaystyle P_{1}:\ {\vec {f}}_{1}t_{1}+{\vec {f}}_{2}{\tfrac {1}{t_{1}}},\ P_{2}:\ {\vec {f}}_{1}t_{2}+{\vec {f}}_{2}{\tfrac {1}{t_{2}}}}$ the points

$${\displaystyle A:\ {\vec {a}}={\vec {f}}_{1}t_{1}+{\vec {f}}_{2}{\tfrac {1}{t_{2}}},\ B:\ {\vec {b}}={\vec {f}}_{1}t_{2}+{\vec {f}}_{2}{\tfrac {1}{t_{1}}}}$$

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

The simple proof is a consequence of the equation ${\displaystyle {\tfrac {1}{t_{1}}}{\vec {a}}={\tfrac {1}{t_{2}}}{\vec {b}}}$.

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.[14]

Hyperbola: tangent-asymptotes-triangle

For simplicity the center of the hyperbola may be the origin and the vectors ${\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}}$ 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 ${\displaystyle \pm ({\vec {f}}_{1}+{\vec {f}}_{2})}$ are the vertices, ${\displaystyle \pm ({\vec {f}}_{1}-{\vec {f}}_{2})}$ span the minor axis and one gets ${\displaystyle |{\vec {f}}_{1}+{\vec {f}}_{2}|=a}$ and ${\displaystyle |{\vec {f}}_{1}-{\vec {f}}_{2}|=b}$.

For the intersection points of the tangent at point ${\displaystyle {\vec {p}}(t_{0})={\vec {f}}_{1}t_{0}+{\vec {f}}_{2}{\tfrac {1}{t_{0}}}}$ with the asymptotes one gets the points

$${\displaystyle C=2t_{0}{\vec {f}}_{1},\ D={\tfrac {2}{t_{0}}}{\vec {f}}_{2}.}$$

The area of the triangle ${\displaystyle M,C,D}$ can be calculated by a 2 × 2 determinant:

$${\displaystyle A={\tfrac {1}{2}}{\Big |}\det \left(2t_{0}{\vec {f}}_{1},{\tfrac {2}{t_{0}}}{\vec {f}}_{2}\right){\Big |}=2{\Big |}\det \left({\vec {f}}_{1},{\vec {f}}_{2}\right){\Big |}}$$

(see rules for determinants). ${\displaystyle \left|\det({\vec {f}}_{1},{\vec {f}}_{2})\right|}$ is the area of the rhombus generated by ${\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}}$. The area of a rhombus is equal to one half of the product of its diagonals. The diagonals are the semi-axes ${\displaystyle a,b}$ of the hyperbola. Hence:

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

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.

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

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

This determinant is conventionally called the discriminant of the conic section.[15]

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

$${\displaystyle \Delta$$ :={\begin{vmatrix}A_{xx}&A_{xy}&B_{x}\\A_{xy}&A_{yy}&B_{y}\\B_{x}&B_{y}&C\end{vmatrix}}=0.}

This determinant ${\displaystyle \Delta }$ is sometimes called the discriminant of the conic section.[16]

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:

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

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 ${\displaystyle (x,y)}$:

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

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

$${\displaystyle {\begin{aligned}x_{c}&=-{\frac {1}{D}}\,{\begin{vmatrix}B_{x}&A_{xy}\\B_{y}&A_{yy}\end{vmatrix}}\,,\\[1ex]y_{c}&=-{\frac {1}{D}}\,{\begin{vmatrix}A_{xx}&B_{x}\\A_{xy}&B_{y}\end{vmatrix}}\,.\end{aligned}}}$$

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

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

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

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

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

$${\displaystyle {\begin{aligned}a^{2}&=-{\frac {\Delta }{\lambda _{1}D}}=-{\frac {\Delta }{\lambda _{1}^{2}\lambda _{2}}},\\[1ex]b^{2}&=-{\frac {\Delta }{\lambda _{2}D}}=-{\frac {\Delta }{\lambda _{1}\lambda _{2}^{2}}},\end{aligned}}}$$

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

$${\displaystyle {\begin{aligned}E&=A_{xx}x_{0}+A_{xy}y_{0}+B_{x},\\[1ex]F&=A_{xy}x_{0}+A_{yy}y_{0}+B_{y},\\[1ex]G&=B_{x}x_{0}+B_{y}y_{0}+C.\end{aligned}}}$$

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

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

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