Question

If the point $P$ on the curve, $4 x^{2}+5 y^{2}=20$ is farthest from the point $Q (0,-4),$ then $PQ ^{2}$ is equal to :

• A. 21
• B. 36
• C. 48
• D. 29

Answer 1

Answer: (B)

Given ellipse is $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$
Let point $P$ is $(\sqrt{5} \cos \theta, 2 \sin \theta)$
$( PQ )^{2}=5 \cos ^{2} \theta+4(\sin \theta+2)^{2}$
$( PQ )^{2}=\cos ^{2} \theta+16 \sin \theta+20$
$( PQ )^{2}=-\sin ^{2} \theta+16 \sin \theta+21$
$=85-(\sin \theta-8)^{2}$
will be maximum when $\sin \theta=1$
$\Rightarrow ( PQ )_{\max }^{2}=85-49=36$