Mathematics · 2024
JEE Main · 27 January 2024, Shift 1 · Q10
If the shortest distance of the parabola y^2=4 x from the centre of the circle x^2+y^2-4 x-16 y+64=0 is d, then d^2 is equal to:
If the shortest distance of the parabola $\displaystyle y^2=4 x$ from the centre of the circle $\displaystyle x^2+y^2-4 x-16 y+64=0$ is d , then $\displaystyle \mathrm{d}^2$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 20$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.