Question

Write the eccentricity of the ellipse
$9x^2+5y^2-18x-2y-16=0.$

Answer 1

Given: Equation of ellipse,

$9x^2+5y^2-18x-2y-16=0.$

$\Rightarrow 9(x^2-2x+1)+5(y^2-\frac{2y}{5}+\frac{1}{25})-9-\frac{1}{5}-16=0$

$\Rightarrow 9(x-1{)}^2+5{(y-\frac{1}{5})}^2=\frac{126}{5}$

$\Rightarrow \frac{9(x-1{)}^2}{\frac{126}{5}}+\frac{5{(y-\frac{1}{5})}^2}{\frac{126}{5}}=1$

$\Rightarrow \frac{(x-1{)}^2}{\frac{126}{45}}+\frac{{(y-\frac{1}{5})}^2}{\frac{126}{25}}=1$

On comparing with standard form,

$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1,$

we get, $a^2=\frac{126}{45}and{b}^2=\frac{126}{25}$

Eccentricity,

$e=\sqrt{1-\frac{a^2}{b^2}}[\because b>a]$


$e=\sqrt{1-\frac{\frac{126}{45}}{\frac{126}{25}}}$

$e=\sqrt{1-\frac{25}{45}}$

$e=\sqrt{1-\frac{5}{9}}=\sqrt{\frac{4}{9}}$

$\Rightarrow e=\frac{2}{3}$