Question

The equation of the ellipse whose foci are$(\pm2,0)$ and eccentricity $\frac{1}{2}$ is

• A. $\frac{x^2}{12}+\frac{y^2}{16}=1$
• B. $\frac{x^2}{16}+\frac{y^2}{12}=1$
• C. $\frac{x^2}{16}+\frac{y^2}{8}=1$
• D. none of these

Answer 1

Answer: (B)

Let the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 $
Given $e= \frac{1}{2}$ and $ae = 2$
$\therefore a=4 $
Since $b^{2} = a^{2}\left(1-e^{2}\right) $
$ = 16\left(1-\frac{1}{4}\right) = \frac{16\times3}{4} = 12 $
$\therefore $ required ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{12} =1$