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:

Question

Accepted Answer

Correct answer: D. (d) JEE Main 2020 Curves is C ≡ x^25+ y^24=1 Let a point on curve be ( 5 cos ,2 sin ) PQ^2= ( 5 cos )^2+(-4-2 sin ) =5 cos^2 +4 sin^2 +16+16 sin ⇒PQ^2=21+16 sin - sin^2 =21+64-( sin -8)^2 =85-( sin -8)^2 For PQ^2 to be maximum sin =1 PQ_ max^2=85-49=36

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

Solution

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