Pages

Subscribe:

Labels

Friday, 5 October 2012

Quadratic equation of Hyperbola

A hyperbola can also be defined as a second-degree equation in the Cartesian coordinates (x, y) of 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 Axx, Axy, Ayy, Bx, By, and C satisfy the determinant condition

D = \begin{vmatrix} A_{xx} & A_{xy}\\A_{xy} & A_{yy} \end{vmatrix} < 0\,
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 Δ is sometimes called the discriminant of the conic section.

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 parameters of the quadratic form.

The center (xc, yc) of the hyperbola may be determined from the formulae

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}
In terms of new coordinates, ξ = xxc and η = yyc, 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 Φ with the positive x-axis that equals

\tan 2\Phi = \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

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}}
where λ1 and λ2 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 is

\frac{{x}^{2}}{a^{2}} - \frac{{y}^{2}}{b^{2}} = 0
The tangent line to a given point (x0, y0) on the hyperbola is defined by the equation

E x + F y + G = 0
where E, F and G are defined

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

F \left( x - x_{0} \right) - E \left( y - y_{0} \right) = 0
The normal line is perpendicular to the tangent line, and both pass through the same point (x0, y0).
From the equation
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \qquad 0 < b \leq a
the basic property that with  r_1 \,\! and  r_2 \,\! being the distances from a point  (x,y) \,\! to the left focus  (-a e , 0) \,\! and the right focus  (a e , 0) \,\! one has for a point on the right branch that
 r_1 - r_2 =2 a\,\!
and for a point on the left branch that
 r_2 - r_1 =2 a\,\!
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 + (x^2-a^2)(e^2-1)=
(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 + (x^2-a^2)(e^2-1)=
(e x - a)^2
If x,y is a point on the right branch of the hyperbola then e x > a\,\! and
 r_1 =e x + a\,\!
 r_2 =e x - a\,\!
Subtracting these equations one gets
 r_1 - r_2 =2 a\,\!
If x,y is a point on the left branch of the hyperbola then e x < -a\,\! and
 r_1 = -e x - a\,\!
 r_2 = -e x + a\,\!
Subtracting these equations one gets
 r_2 - r_1 =2 a\,\!

0 comments:

Post a Comment