The eccentricity of an ellipse having center at the origin, axes along the co-ordinate axes and passing

through the points $(4,-1)$ and $(-2,2)$ is

Let equation of the ellipse be

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Ellipse passes through the points $(4,-1)$ and $(-2,2)$

$\therefore$ $\frac{16}{a^2}+\frac{1}{b^2}=1$

$\Rightarrow 16b^2+a^2=a^2{b}^2$                                           $\dots \left(1\right)$

And $\frac{4}{a^2}+\frac{4}{b^2}=1$

$\Rightarrow 4b^2+4a^2=a^2{b}^2$

From $\text{ (1) \& (2), }$we get

$\begin{array}{l}16b^2+a^2=4a^2+4b^2\\ \Rightarrow 3a^2=12b^2\Rightarrow a^2=4b^2\end{array}$

Now, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{b^2}{4b^2}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$