Question

The eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ if its latus rectum is equal to one half of its minor axis, is
(a) $\frac{1}{\sqrt{2}}$

(b) $\frac{\sqrt{3}}{2}$

(c) $\frac{1}{2}$

(d) none of these

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)e=\frac{\sqrt{3}}{2} \\ \text{According to the question, the latus rectum is half its minor axis.} \\ \mathrm{i}.\mathrm{e}.\frac{2b^2}{a}=\frac{1}{2}\times 2b \\ \Rightarrow 2b^2=ab \\ \Rightarrow a=2b \\ \mathrm{Now},e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow e=\sqrt{1-\frac{b^2}{4b^2}} \\ \Rightarrow e=\sqrt{1-\frac{1}{4}} \\ \Rightarrow e=\frac{\sqrt{3}}{2} \end{gathered}$$