# The eccentricity of the conic $9x^{2}+5y^{2}-54x-40y+116=0$ is

$\begin{array}{1 1}(1)\frac{1}{3}&(2)\frac{2}{3}\\(3)\frac{4}{9}&(4)\frac{2}{\sqrt{5}}\end{array}$

$9x^2+5y^2-54x-40y+116=0$
$9x^2+54x+5y^2-40y+116=0$
$9[x^2-6x]+5[y^2-8y]+116=0$
$9[(x-3)^2-3^2]+5[(y-4)^2-4^2]+116=0$
$9(x-3)^2-81+5(y-4)^2-80+116=0$
$9(x-3)^2+5(y-4)^2-45=0$
$9(x-3)^2+5(y-4)^2=45$
$\large\frac{9(x-3)^2}{5} +\frac{5 (y-4)^2}{45}$$=1 \large\frac{(x-3)^2}{5} +\frac {(y-4)}{9}$$=1$
$a^2=9,b^2=5$
$e= \sqrt {1-\large\frac{b^2}{a^2}}$
$\qquad= \sqrt {1- \large\frac{5}{9}}$
$\qquad=\sqrt {\large\frac{4}{9}}$
$e= \large\frac{2}{3}$
Hence 2 is the correct answer.