Question

In an ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where a < b, eccentricity 'e' is given by

Answer 1

In ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a<b$

$a^2=b^2(1-e^2)$

$\frac{a^2}{b^2}=1-e^2$

$e^2=1-\frac{a^2}{b^2}$

$e=\sqrt{1-\frac{a^2}{b^2}}$