Question

The foci of an ellipse are $(0,\pm 2)$ and its eccentricity is $\frac{1}{\sqrt{2}}.$ Find its equation.

• A. $\frac{x^2}{4}-\frac{y^2}{8}=1$
• B. $\frac{x^2}{8}+\frac{y^2}{4}=1$
• C. $\frac{x^2}{4}+\frac{y^2}{8}=1$
• D. $\frac{x^2}{8}-\frac{y^2}{4}=1$

Answer 1

Answer: (C)

$\frac{x^2}{4}+\frac{y^2}{8}=1$

Foci are given to be $(0,\pm 2)$ and eccentricity, $e=\frac{1}{\sqrt{2}}$

Since the foci are on y axes, this is a case of vertical form of ellipse.

In vertical form of ellipse foci is given by $e=(0,\pm \sqrt{b^2-a^2})$

Thus $2=\left(\sqrt{b^2-a^2}\right)\Rightarrow b^2=a^2+4$ ........(1)

Now eccentricity $e=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{a^2}{b^2}}$

Thus $b^2=2a^2$ ......(2)

Solving 1) and 2) we get $a=2,b=2\sqrt{2}$

Thus equation of the ellipse is $\frac{x^2}{4}+\frac{y^2}{8}=1.$