Q79 / 200 · Manipal (MET) 2014 · Mathematics
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
- Correct answerA$\sqrt{5}$
- B$3$
- C$\sqrt{12}$
- D$\frac{7}{2}$
Solution
Given equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{\frac{7}{4}}=1$
Here,
$a^2=4$
and
$b^2=\frac{7}{4}$
$\because$
$b^2=a^2\left(1-e^2\right)$
$\therefore \quad \frac{7}{4}=4\left(1-e^2\right)$
$\Rightarrow \quad e^2=1-\frac{7}{16}=\frac{9}{16} \Rightarrow e=\frac{3}{4}$,
Thus, the foci are $\left( \pm \frac{3}{2}, 0\right)$.
The radius of the required circle
$\begin{aligned}
& =\sqrt{\left(\frac{3}{2}-\frac{1}{2}\right)^2+(0-2)^2} \\
& =\sqrt{5}
\end{aligned}$
Hence, the correct answer is option: 'A'.