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. $48$
• B. $29$
• C. $21$
• D. $36$

Answer 1

The correct option is D $36$

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