Hyperbola
Quadratic equation

A hyperbola can also be defined as a second-degree equation in the Cartesian coordinates (x, y) {\displaystyle (x,y)} in the plane,

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

provided that the constants A x x, {\displaystyle A_{xx},} A x y, {\displaystyle A_{xy},} A y y, {\displaystyle A_{yy},} B x, {\displaystyle B_{x},} B y, {\displaystyle B_{y},} and C {\displaystyle C} satisfy the determinant condition

D:= | A x x A x y A x y A y y | < 0. {\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.

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

Δ:= | A x x A x y B x A x y A y y B y B x B y C | = 0. {\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 Δ {\displaystyle \Delta } is sometimes called the discriminant of the conic section.

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

A x x = − a 2 sin 2 θ + b 2 cos 2 θ, B x = − A x x x ∘ − A x y y ∘, A y y = − a 2 cos 2 θ + b 2 sin 2 θ, B y = − A x y x ∘ − A y y y ∘, A x y = (a 2 + b 2) sin θ cos θ, C = A x x x ∘ 2 + 2 A x y x ∘ y ∘ + A y y y ∘ 2 − a 2 b 2. {\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 },\\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{aligned}}}

These expressions can be derived from the canonical equation

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

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

X = + (x − x ∘) cos θ + (y − y ∘) sin θ, Y = − (x − x ∘) sin θ + (y − y ∘) cos θ. {\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 (x c, y c) {\displaystyle (x_{c},y_{c})} of the hyperbola may be determined from the formulae

x c = − 1 D | B x A x y B y A y y |, y c = − 1 D | A x x B x A x y B y |. {\displaystyle {\begin{aligned}x_{c}&=-{\frac {1}{D}}\,{\begin{vmatrix}B_{x}&A_{xy}\\B_{y}&A_{yy}\end{vmatrix}}\,,\\y_{c}&=-{\frac {1}{D}}\,{\begin{vmatrix}A_{xx}&B_{x}\\A_{xy}&B_{y}\end{vmatrix}}\,.\end{aligned}}}

In terms of new coordinates, ξ = x − x c {\displaystyle \xi =x-x_{c}} and η = y − y c, {\displaystyle \eta =y-y_{c},} the defining equation of the hyperbola can be written

A x x ξ 2 + 2 A x y ξ η + A y y η 2 + Δ D = 0. {\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 φ {\displaystyle \varphi } with the positive x {\displaystyle x} -axis that is given by

tan (2 φ) = 2 A x y A x x − A y y. {\displaystyle \tan(2\varphi)={\frac {2A_{xy}}{A_{xx}-A_{yy}}}.}

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

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

The major and minor semiaxes a {\displaystyle a} and b {\displaystyle b} are defined by the equations

a 2 = − Δ λ 1 D = − Δ λ 1 2 λ 2, b 2 = − Δ λ 2 D = − Δ λ 1 λ 2 2, {\displaystyle {\begin{aligned}a^{2}&=-{\frac {\Delta }{\lambda _{1}D}}=-{\frac {\Delta }{\lambda _{1}^{2}\lambda _{2}}},\\b^{2}&=-{\frac {\Delta }{\lambda _{2}D}}=-{\frac {\Delta }{\lambda _{1}\lambda _{2}^{2}}},\end{aligned}}}

where λ 1 {\displaystyle \lambda _{1}} and λ 2 {\displaystyle \lambda _{2}} are the roots of the quadratic equation

λ 2 − (A x x + A y y) λ + D = 0. {\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

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

The tangent line to a given point (x 0, y 0) {\displaystyle (x_{0},y_{0})} on the hyperbola is defined by the equation

E x + F y + G = 0 {\displaystyle Ex+Fy+G=0}

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

E = A x x x 0 + A x y y 0 + B x, F = A x y x 0 + A y y y 0 + B y, G = B x x 0 + B y y 0 + C. {\displaystyle {\begin{aligned}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{aligned}}}

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

F (x − x 0) − E (y − y 0) = 0. {\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 (x 0, y 0). {\displaystyle (x_{0},y_{0}).}

From the equation

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

the left focus is (− a e, 0) {\displaystyle (-ae,0)} and the right focus is (a e, 0), {\displaystyle (ae,0),} where e {\displaystyle e} is the eccentricity. Denote the distances from a point (x, y) {\displaystyle (x,y)} to the left and right foci as r 1 {\displaystyle r_{1}} and r 2. {\displaystyle r_{2}.} For a point on the right branch,

r 1 − r 2 = 2 a, {\displaystyle r_{1}-r_{2}=2a,}

and for a point on the left branch,

r 2 − r 1 = 2 a. {\displaystyle r_{2}-r_{1}=2a.}

This can be proved as follows:

If (x, y) {\displaystyle (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 + (x 2 − a 2) (e 2 − 1) = (e x + a) 2. {\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

r 2 2 = (x − a e) 2 + y 2 = x 2 − 2 x a e + a 2 e 2 + (x 2 − a 2) (e 2 − 1) = (e x − a) 2. {\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 (x, y) {\displaystyle (x,y)} is a point on the right branch of the hyperbola then e x > a {\displaystyle ex>a} and

r 1 = e x + a, r 2 = e x − a. {\displaystyle {\begin{aligned}r_{1}&=ex+a,\\r_{2}&=ex-a.\end{aligned}}}

Subtracting these equations one gets

r 1 − r 2 = 2 a. {\displaystyle r_{1}-r_{2}=2a.}

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

r 1 = − e x − a, r 2 = − e x + a. {\displaystyle {\begin{aligned}r_{1}&=-ex-a,\\r_{2}&=-ex+a.\end{aligned}}}

Subtracting these equations one gets

r 2 − r 1 = 2 a. {\displaystyle r_{2}-r_{1}=2a.}
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