The distance between the focii of the ellipse $x=3 \cos \theta$, $y=4 \sin \theta$ is
The distance between the focii of the ellipse $x=3 \cos \theta$, $y=4 \sin \theta$ is
$2 \sqrt{7}$
$7 \sqrt{2}$
$\sqrt{7}$
$3 \sqrt{7}$
Solution
Given that, $x=3 \cos \theta$
$\Rightarrow \quad \frac{x}{3}=\cos \theta$
and $y=4 \sin \theta$
$\Rightarrow \quad \frac{y}{4}=\sin \theta$
On squaring and adding eqs. (i) and (ii) we get
$\begin{aligned}
& \left(\frac{x}{3}\right)^2+\left(\frac{y}{4}\right)^2=\cos ^2 \theta+\sin ^2 \theta \\
& \frac{x^2}{9}+\frac{y^2}{16}=1
\end{aligned}$
$\therefore$ Distance between their foci,
$f_1 f_2=2 \sqrt{b^2-a^2}=2 \sqrt{16-9}=2 \sqrt{7}$