QJEEM20Ma92ef: JEE Main 2020 Mathematics - Class 12 JEEM20Ma92ef Solution: If the point P on the curve, 4x 2 + 5y 2 = 20 is farthest from…

Question

Verified Solution • 2025–26 Session moderate

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

Accepted Answer

Given ellipse is

{{{x^2}} \over 5} + {{{y^2}} \over 4} = 1

Let point P is

(\sqrt 5 \cos \theta ,\,2\sin \theta )

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

\Rightarrow

(PQ)² =

5{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16

\Rightarrow

(PQ)² =

{\cos ^2}\theta + 4{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16

\Rightarrow

(PQ)² =

{\cos ^2}\theta + 4 + 16\sin \theta + 16

\Rightarrow

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

\Rightarrow

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

=

- \left( {{{\sin }^2}\theta - 2.8\sin \theta + 64} \right) + 64 + 21

=

85 - {(\sin \theta - 8)^2}

will be maximum when sin

\theta

= 1

\Rightarrow {(PQ)^2}_{\max } = 85 - 49 = 36

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