Question

If e is the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\left(a<b\right),$ then
(a) b² = a²(1 – e²)
(b) a² = b²(1 – e²)
(c) a² = b²(e² – 1)
(d) b² = a²(e² – 1)

Answer 1

For a < b,
Given equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Eccentricity $e=\sqrt{1-\frac{a^2}{b^2}}$
$$\begin{gathered} \mathrm{i}.\mathrm{e}{e}^2=\frac{b^2-a^2}{b^2} \\ \mathrm{i}.\mathrm{e}{b}^2{e}^2=b^2-a^2 \\ \mathrm{i}.\mathrm{e}{a}^2=b^2\left(1-e^2\right) \end{gathered}$$

Hence, the correct answer is option B.