Editing Hyperbola (section)

===Quadratic equation===
A hyperbola can also be defined as a second-degree equation in the Cartesian coordinates <math>(x, y)</math> in the [[plane (geometry)|plane]],

<math display=block>
A_{xx} x^2 + 2 A_{xy} xy + A_{yy} y^2 + 2 B_x x + 2 B_y y + C = 0,
</math>

provided that the constants <math>A_{xx},</math> <math>A_{xy},</math> <math>A_{yy},</math> <math>B_x,</math> <math>B_y,</math> and <math>C</math> satisfy the determinant condition

<math display=block>
D := \begin{vmatrix}
  A_{xx} & A_{xy} \\
  A_{xy} & A_{yy} \end{vmatrix} < 0.
</math>

This determinant is conventionally called the [[discriminant#Discriminant of a conic section|discriminant]] of the conic section.<ref>{{cite book
|title=Math refresher for scientists and engineers
|last1=Fanchi
|first1=John R.
|publisher=John Wiley and Sons
|year=2006
|isbn=0-471-75715-2
|url=https://books.google.com/books?id=75mAJPcAWT8C
|at=[https://books.google.com/books?id=75mAJPcAWT8C&pg=PA44 Section 3.2, pages 44–45]
}}</ref>

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

<math display=block>
\Delta := \begin{vmatrix}
  A_{xx} & A_{xy} & B_x \\
  A_{xy} & A_{yy} & B_y \\
  B_x & B_y & C
\end{vmatrix} = 0.
</math>

This determinant <math>\Delta</math> is sometimes called the discriminant of the conic section.<ref>{{cite book |last1=Korn |first1=Granino A |author2-link=Theresa M. Korn |last2=Korn |first2=Theresa M. |title=Mathematical Handbook for Scientists and Engineers: Definitions, Theorems, and Formulas for Reference and Review |publisher=Dover Publ. |edition=second |year=2000 |page=40}}</ref>

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

<math display=block>\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}</math>

These expressions can be derived from the canonical equation

<math display=block>\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1</math>

by a [[rigid transformation|translation and rotation]] of the coordinates {{nobr|<math>(x, y)</math>:}}

<math display=block>\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}</math>

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 <math>(x_c, y_c)</math> of the hyperbola may be determined from the formulae

<math display=block>\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}</math>

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

<math display=block>
A_{xx} \xi^2 + 2A_{xy} \xi\eta + A_{yy} \eta^2 + \frac \Delta D = 0.
</math>

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

<math display=block>\tan (2\varphi) = \frac{2A_{xy}}{A_{xx} - A_{yy}}.</math>

Rotating the coordinate axes so that the <math>x</math>-axis is aligned with the transverse axis brings the equation into its '''canonical form'''

<math display=block>\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.</math>

The major and minor semiaxes <math>a</math> and <math>b</math> are defined by the equations

<math display=block>\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}</math>

where <math>\lambda_1</math> and <math>\lambda_2</math> are the [[root of a function|roots]] of the [[quadratic equation]]

<math display=block>\lambda^2 - \left( A_{xx} + A_{yy} \right)\lambda + D = 0.</math>

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

<math display=block>\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0.</math>

The tangent line to a given point <math>(x_0, y_0)</math> on the hyperbola is defined by the equation

<math display=block>E x + F y + G = 0</math>

where <math>E,</math> <math>F,</math> and <math>G</math> are defined by

<math display=block>\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}</math>

The [[normal (geometry)|normal line]] to the hyperbola at the same point is given by the equation

<math display=block>F(x - x_0) - E(y - y_0) = 0.</math>

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

From the equation

<math display=block>\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \qquad 0 < b \leq a,</math>

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

<math display=block> r_1 - r_2 = 2 a, </math>

and for a point on the left branch,

<math display=block> r_2 - r_1 = 2 a. </math>

This can be proved as follows:

If <math>(x, y)</math> is a point on the hyperbola the distance to the left focal point is

<math display=block>
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.
</math>

To the right focal point the distance is

<math display=block>
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.
</math>

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

<math display=block>\begin{align}
r_1 &= e x + a, \\
r_2 &= e x - a.
\end{align}</math>

Subtracting these equations one gets

<math display=block>r_1 - r_2 = 2a.</math>

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

<math display=block>\begin{align}
r_1 &= - e x - a, \\
r_2 &= - e x + a.
\end{align}</math>

Subtracting these equations one gets

<math display=block>r_2 - r_1 = 2a.</math>