If the point P on the curve 4x^{2} + 5y^{2} = | vedclass

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(B) The given equation of the ellipse is $4x^{2} + 5y^{2} = 20$,which can be written as $\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1$.Let the point $P$ on t

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) The given equation of the ellipse is $4x^{2} + 5y^{2} = 20$,which can be written as $\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1$.Let the point $P$ on the ellipse be $(\sqrt{5} \cos 	heta, 2 \sin 	heta)$.The square of the distance $PQ$ is given by $PQ^{2} = (\sqrt{5} \cos 	heta - 0)^{2} + (2 \sin 	heta - (-4))^{2}$.$PQ^{2} = 5 \cos^{2} 	heta + (2 \sin 	heta + 4)^{2}$.$PQ^{2} = 5(1 - \sin^{2} 	heta) + 4 \sin^{2} 	heta + 16 \sin 	heta + 16$.$PQ^{2} = 5 - 5 \sin^{2} 	heta + 4 \sin^{2} 	heta + 16 \sin 	heta + 16$.$PQ^{2} = -\sin^{2} 	heta + 16 \sin 	heta + 21$.To find the maximum value,we complete the square: $PQ^{2} = -(\sin^{2} 	heta - 16 \sin 	heta + 64) + 64 + 21$.$PQ^{2} = 85 - (\sin 	heta - 8)^{2}$.Since $-1 \leq \sin 	heta \leq 1$,the expression $85 - (\sin 	heta - 8)^{2}$ is maximized when $\sin 	heta$ is as close to $8$ as possible,which occurs at $\sin 	heta = 1$.Substituting $\sin 	heta = 1$,we get $PQ^{2} = 85 - (1 - 8)^{2} = 85 - (-7)^{2} = 85 - 49 = 36$.

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

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