Question

There are exactly two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ whose distance from its center 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

$\frac{1}{\sqrt{2}}$

Explanation for correct option:

Eccentricity of the ellipse:

As we know that 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. $\sqrt{\frac{a^2+2b^2}{2}}>b$, so the points must be the vertices of major axis.

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

Squaring on both sides, we get,

$\begin{array}{rcl}{a}^2& =& \frac{a^2+2b^2}{2}\\ 2a^2& =& a^2+2b^2\\ a^2& =& 2b^2\end{array}$

Now, eccentricity of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

$$\begin{gathered} =\sqrt{1-\frac{b^2}{a^2}} \\ =\sqrt{1-\frac{b^2}{2b^2}} \\ =\sqrt{1-\frac{1}{2}} \\ =\frac{1}{\sqrt{2}} \end{gathered}$$

Hence, Option B = $\frac{1}{\sqrt{2}}$ is the correct answer.