Question

If for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ the length of latus-rectum is half of the minor axis. Then the eccentricity of the ellipse is

• A. $\frac{\sqrt{2}}{3}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{4}$
• D. none of these

Answer 1

Answer: (B)

$\frac{\sqrt{3}}{2}$

Given the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$......(1).

Now length of the latus-rectum is $\frac{2b^2}{a}$ and the length of the minor axis is $2b$.

According to the problem

$\frac{2b^2}{a}=b$

or, $b=\frac{a}{2}$.....(2).

Now eccentricity of the ellipse if $\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$ [ Using (2)].