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 are the same and are equal to $\sqrt{a^2+2b^2}/2$. Then the eccentricity of the ellipse is

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

Answer 1

Answer: (B)

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 the same, the points would either be the endpoints of the major axis or of the minor axis.
But $\sqrt{\left(a^2+2b^2\right)/2}>$ b.
So, the points are the vertices of the major axis.
Hence, $a=\sqrt{\frac{a^2+2b^2}{2}}$
or $a^2=2b^2$
or $e=\sqrt{1-\frac{b^2}{a^2}}=\frac{1}{\sqrt{2}}$