MCQ
The eccentricity of the ellipse $\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{y}^2}=1$ if its latus rectum is equal to one half of its minor axis, is:
  • A
    $\frac{1}{\sqrt{2}}$
  • $\frac{\sqrt{3}}{2}$
  • C
    $\frac{1}{2}$
  • D
    $\text{none of these}$

Answer

Correct option: B.
$\frac{\sqrt{3}}{2}$
According to the question, the latus rectum is half its minor axis.
i.e. $\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{b}$
$\Rightarrow2\text{b}^2=\text{ab}$
$\Rightarrow\text{a}=2\text{b}$
Now, $\text{e}\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{\text{b}^2}{4\text{b}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{1}{4}}$
$\Rightarrow\text{e}=\frac{\sqrt{3}}{2}$

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