Question

The equation of the ellipse whose foci are$(\pm 2,0)$ 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 $b^2=a^2\left(1-e^2\right)$
$=16\left(1-\frac{1}{4}\right)=\frac{16\times 3}{4}=12$
$\therefore$ required ellipse is $\frac{x^2}{16}+\frac{y^2}{12}=1$