Question

Find the eccentricity and length of latus rectum of the ellipse $9x^2+16y^2-36x+32y-92=0$.

Answer 1

We have $9x^2+16y^2-36x+32y-92=0$

$\Rightarrow 9(x^2-4x)+16(y^2+2y)=92$

$\Rightarrow 9(x^2-4x+4)+16(y^2+2y+1)=92+36+16=144$

$\Rightarrow 9(x-2{)}^2+16(y+1{)}^2=144$

$\Rightarrow \frac{(x-2{)}^2}{16}+\frac{(y+1{)}^2}{9}=1$

$\Rightarrow a^2=16,b^2=9$

Thus eccentricity, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

and length of latus rectum $=\frac{2b^2}{a}=\frac{9}{2}$