If the point P on the curve, 4x² + 5y² = 20 is
farthest from the point Q(0, -4), then PQ² is equal to:

Given ellipse is ${{{x^2}} \over 5} + {{{y^2}} \over 4} = 1$<br><br>Let point P is $(\sqrt 5 \cos \theta ,\,2\sin \theta )$<br><br>${(PQ)^2}=5{\cos ^2}\theta + {(2\sin \theta + 4)^2}$ <br><br>$\Rightarrow$ (PQ)² = $5{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16$ <br><br>$\Rightarrow$ (PQ)² = ${\cos ^2}\theta + 4{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16$ <br><br>$\Rightarrow$ (PQ)² = ${\cos ^2}\theta + 4 + 16\sin \theta + 16$ <br><br>$\Rightarrow$ ${(PQ)^2} = {\cos ^2}\theta + 16\sin \theta + 20$ <br><br>$\Rightarrow$ ${(PQ)^2} = - {\sin ^2}\theta + 16\sin \theta + 21$ <br><br>= $- \left( {{{\sin }^2}\theta - 2.8\sin \theta + 64} \right) + 64 + 21$ <br><br>= $85 - {(\sin \theta - 8)^2}$<br><br>will be maximum when sin $\theta$ = 1<br><br>$\Rightarrow {(PQ)^2}_{\max } = 85 - 49 = 36$