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

The correct option is B $\frac{\sqrt{3}}{2}$


According to the question, the latus rectum is half its minor axis
i.e. $\frac{2b^2}{a}=\frac{1}{2}\times 2b$
$\Rightarrow 2b^2=ab$
$\Rightarrow a=2a$
Now, $e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow e=\sqrt{1-\frac{b^2}{4b^2}}$
$\Rightarrow e=\sqrt{1-\frac{1}{4}}$
$\Rightarrow e=\frac{\sqrt{3}}{2}$