Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$ Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$
Solution : 
$4(2y-x-3)^2 = 4x^2-16xy+24x+16y^2-48y+36$ 
and $9(2x+y-1)^2 = 36x^2+36xy-36x+9y^2-18y+9$
$\therefore 4(2y-x-3)^2 -9(2x+y-1)^2 = 7y^2+60x -52xy-32x^2-30y+27 =80$
Can we have other option available so that we will be able to find the solution more quicker way,
Since this is a conic then the  given lines let $L_1 =2y-x-3=0 $ and $L_2 = 2x+y-1=0$ are perpendicular to each other .. Can we use this somehow please suggest... thanks..

With a translation of the origin in the center of the conic (which is the point of intersection of the two lines, you get the equation in the form
$$
4(-X+2Y)^2-9(2X+Y)^2=80
$$
Now the lines are indeed orthogonal: since $1^2+2^2=5$, you can write the equation as
$$
4\cdot5\left(-\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y\right)^2
-9\cdot5\left(\frac{2}{\sqrt{5}}X+\frac{1}{\sqrt{5}}Y\right)^2=80
$$
The transformation
\begin{cases}
\xi=-\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y\\
\eta=\frac{2}{\sqrt{5}}X+\frac{1}{\sqrt{5}}Y
\end{cases}
is a rotation, so you get the canonical form
$$
4\xi^2-9\eta^2=16
$$
or
$$
\frac{\xi^2}{4}-\frac{\eta^2}{16/9}=1
$$
Thus you have
$$a^2=4,\quad b^2=\frac{16}{9}$$
