# There are exactly two points on the ellipse x^2/(a^2)+y^2/(b^2)=1 whose distance from its centre is same and equal to (1/sqrt2)(sqrt(a^2+2b^2)). The eccentricity of the ellipse is: A. 1/2 B. 1/√2 C. 3√2 D.1/3√2

As per the given condition,

a = √[(a2 +2b2)/√2]

Squareing both the sides we get;

a2 = 2b2

b2/a2 = 1/2

By eccentricity formula,

$$Eccentricity, \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}} = \sqrt{1-1/2} = \sqrt{1/2}$$