Mathematics · 2020
JEE Main · 5 September 2020, Shift 1 · Q60
If the point P on the curve, 4 x^2+5 y^2=20 is farthest from the point Q (0,-4), then PQ^2 is equal to:
If the point P on the curve, $\displaystyle 4 x^2+5 y^2=20$ is farthest from the point $\displaystyle \mathrm{Q}(0,-4)$, then $\displaystyle \mathrm{PQ}^2$ is equal to :
Official answer
(1)
$\displaystyle 36$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.