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