MCQ
Choose the correct answer. If $e$ is the eccentricity of the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1(\text{a}<\text{b}),$ then:
  • A
    $ b^2=a^2\left(1-e^2\right) $
  • $ a^2=b^2\left(1-e^2\right) $
  • C
    $ a^2=b^2\left(e^2-1\right) $
  • D
    $ b^2=a^2\left(e^2-1\right) $

Answer

Correct option: B.
$ a^2=b^2\left(1-e^2\right) $
Given equation is $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1(\text{a}<\text{b})$
$\therefore\text{ Eccentricity e}=\sqrt{1-\frac{\text{a}^2}{\text{b}^2}}$
$\Rightarrow\text{e}^2=1-\frac{\text{a}^2}{\text{b}^2}$
$\Rightarrow\frac{\text{a}^2}{\text{b}^2}=(1-\text{e}^2)$
$\Rightarrow\text{a}^2=\text{b}^2(1-\text{e}^2)$

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