Question

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

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

Answer 1

Answer: (B)

$\because a e= \pm \sqrt{5}$
$\Rightarrow a= \pm \sqrt{5}\left(\frac{3}{\sqrt{5}}\right)= \pm 3$
$\Rightarrow a^2=9$
$\therefore b^2=a^2\left(1-e^2\right)=9\left(1-\frac{5}{9}\right)=4$
Hence, equation of ellipse
$\frac{x^2}{9}+\frac{y^2}{4}=1$
$\Rightarrow 4 x^2+9 y^2=36$