Question

If the point $P$ on the curve, $4x^2+5y^2=20$ is farthest from the point $Q(0,-4),$ then $P{Q}^2$ 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 $\sin \theta =1$
$\Rightarrow (PQ{)}_{\max }^2=85-49=36$