Question

There are exactly two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ whose distance from its centre is the same and is equal to $\sqrt{\frac{a^2+2b^2}{2}}$. Then the eccentricity of the ellipse is:

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{3}$
• D. $\frac{1}{3\sqrt{2}}$

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}$

Since there are exactly two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, whose distance from the center is same, the points would be either endpoints of the major axis or of the minor axis.

But $\frac{\sqrt{a^2+2b^2}}{\sqrt{2}}>b$, so the points are the vertices of major axis.

Hence, $a=\frac{\sqrt{a^2+2b^2}}{\sqrt{2}}$

Squaring on both the side we get,

$\Rightarrow a^2=2b^2$

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