Question

If the foci of an ellipse are $(\pm \sqrt{5},0)$ and its eccentricity is $\sqrt{5}/3,$ then the equation of the ellipse is

• A. $9x^2+4y^2=36$
• B. $4x^2+9y^2=36$
• C. $36x^2+9y^2=4$
• D. $9x^2+36y^2=4$

Answer 1

Answer: (B)

Given, eccentricity $e=\frac{\sqrt{5}}{3}$
and foci $=(\pm \sqrt{5},0)$
$\Rightarrow$ $ae=\sqrt{5}$
$\Rightarrow$ $a=\frac{\sqrt{5}\times 3}{\sqrt{5}}=3$
Now, $b^2=a^2(1-e^2)=9\left(1-\frac{5}{9}\right)$
$\Rightarrow$ $b^2=4$
$\therefore$ The equation of ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow$ $\frac{x^2}{9}+\frac{y^2}{4}=1$
$\Rightarrow$ $4x^2+9y^2=36$