Thank you for reporting, we will resolve it shortly
Q.
The radius of the circle passing through the foci of the ellipse $\frac{x^{2}}{4}+\frac{4 y^{2}}{7}=1$ and having its centre at $\left(\frac{1}{2}, 2\right)$ is
ManipalManipal 2014
Solution:
Given equation of ellipse is
$\frac{x^{2}}{4}+\frac{y^{2}}{\frac{7}{4}}=1$
Here, $ a^{2}=4$
and $ b^{2}=\frac{7}{4}$
$\because b^{2}=a^{2}\left(1-e^{2}\right)$
$\therefore \frac{7}{4}=4\left(1-e^{2}\right)$
$\Rightarrow e^{2}=1-\frac{7}{16}=\frac{9}{16}$
$\Rightarrow e=\frac{3}{4}$
Thus, the foci are $\left(\pm \frac{3}{2}, 0\right)$.
The radius of the required circle
$=\sqrt{\left(\frac{3}{2}-\frac{1}{2}\right)^{2}+(0-2)^{2}}$
$=\sqrt{5}$