<s>[INST] 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..
 [/INST] As the Rotation of the axes does not change the eccentricity of a curve
set $2y-x-3=X,2x+y−1 =Y$ so that the given equation becomes  $$4X^2-9Y^2=80\iff \frac{X^2}{\frac{80}4}-\frac{Y^2}{\frac{80}9}=1$$
Now we know, $$b^2=a^2(e^2-1)$$ where $a,b\le a$ are the semi-major & semi-minor axes and $e$ is the eccentricity of the hyperbola 
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