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