Question

how to reduce 2xsquared-3ysquare-6x+12y to standard form of a hyperbola

Answer 1

Answer on Question 40061, Math, Analytic Geometry Question: how to reduce 2xsquared-3ysquare-6x+12y to standard form of a hyperbola. Solution. First let us find the discriminant of the conic section $\Delta$. If the equation is

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

Then, it is defined as:

\[ \Delta:=\begin{vmatrix}A_{xx}&A_{xy}&B_{x}\\

A_{xy}&A_{yy}&B_{y}\\

B_{x}&B_{y}&C\end{vmatrix} \]

In our case

\[ \Delta=\begin{vmatrix}2&0&-3\\

0&-3&6\\

-3&6&0\end{vmatrix}=-45 \]

Next we must find roots of quadratic equation

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

where D is determinant

\[ D=\begin{vmatrix}2&0\\

0&-3\end{vmatrix}=-6 \]

$\lambda^{2}-\lambda-6=0$

Roots are $\lambda_{1}=3,\lambda_{2}=-2$ So, the $a$ and $b$ of the canonical form

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

can be found as

$a^{2}=-\frac{\Delta}{\lambda_{1}^{2}\lambda_{2}},\qquad b^{2}=\frac{\Delta}{\lambda_{1}\lambda_{2}^{2}}$

$a=5/2,\quad b=15/4$

The canonical form is

$\frac{x^{2}}{5/2}-\frac{y^{2}}{15/4}=1$