Question

If the point P on the curve, $4x^2+5y^2=20$ is farthest from the point Q$(0,-4)$, then PQ² is equal to:

• A. $48$
• B. $29$
• C. $21$
• D. $36$

Answer 1

The correct option is D

$36$

step1: Given ellipse is $\left(\frac{x^2}{5}\right)+\left(\frac{y^2}{4}\right)=1$

Let point $P$ is $(\sqrt{5}\cos \theta ,2\sin \theta )$

${\left(PQ\right)}^2=5{\cos }^2\theta +4{(\sin \theta +2)}^2$

${\left(PQ\right)}^2={\cos }^2\theta +16\sin \theta +20$

${\left(PQ\right)}^2=-{\sin }^2\theta +16\sin \theta +21$

${\left(PQ\right)}^2=85-{(\sin \theta –8)}^2$

${\left(PQ\right)}^2$ Will be maximum when $\sin \theta =1$

${\left(PQ\right)}^2$max = $85-49=36$

Hence, option $\mathbf{\left(}\mathit{D}\mathbf{\right)}$ is correct.