JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 13)
If the point P on the curve, 4x2 + 5y2 = 20 is
farthest from the point Q(0, -4), then PQ2 is equal to:
farthest from the point Q(0, -4), then PQ2 is equal to:
36
48
21
29
Explanation
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
$$ \Rightarrow {(PQ)^2}_{\max } = 85 - 49 = 36$$
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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