Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{4}{9}$
• D. $\frac{2}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{2}{3}$

Given conic is $9x^2+5y^2-54x-40y+116=0$

$\Rightarrow 9(x^2-6x)+5(y^2-8y)+116=0$

$\Rightarrow 9(x-3{)}^2+5(y-4{)}^2-81-80+116=0$

$\Rightarrow 9(x-3{)}^2+5(y-4{)}^2-45=0$

$\Rightarrow 9(x-3{)}^2+5(y-4{)}^2=45$

Divide both sides by $45$, we get

$\frac{(x-3{)}^2}{5}+\frac{(y-4{)}^2}{9}=1$ which is in the standard form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ of ellipse.

Thus $b^2=5,a^2=9$

Eccentricity $=\sqrt{1-\frac{b^2}{a^2}}$

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

$=\sqrt{\frac{9-5}{9}}$

$=\sqrt{\frac{4}{9}}$

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