Question:
If the point $P$ on the curve, $4 x^{2}+5 y^{2}=20$ is farthest from the point $\mathrm{Q}(0,-4)$, then $\mathrm{PQ}^{2}$ is equals to :
Correct Option: 1,
Solution:
Ellipse $\equiv \frac{x^{2}}{5}+\frac{y^{2}}{4}=1$
Let a point on ellipse be $(\sqrt{5} \cos \theta, 2 \sin \theta)$
$\therefore P Q^{2}=(\sqrt{5} \cos \theta)^{2}+(-4-2 \sin \theta)^{2}$
$=5 \cos ^{2} \theta+4 \sin ^{2} \theta+16+16 \sin \theta$
$=21+16 \sin \theta-\sin ^{2} \theta$
$=21+64-(\sin \theta-8)^{2}=85-(\sin \theta-8)^{2}$
$P Q^{2}$ to be maximum when $\sin \theta=1$
$\therefore P Q_{\max }^{2}=85-49=36$
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