Mathematics · 2022
JEE Main · 25 June 2022, Shift 1 · Q19
Let x=2 t, y=(t^2)/3 be a conic. Let S be the focus and B be the point on the axis of the conic such that SA ⊥ BA, where A is any point on the conic.…
Let $\displaystyle x=2 \mathrm{t}, y=\frac{\mathrm{t}^2}{3}$ be a conic. Let S be the focus and B be the point on the axis of the conic such that $\displaystyle \mathrm{SA} \perp \mathrm{BA}$, where A is any point on the conic. If k is the ordinate of the centroid of the $\displaystyle \triangle \mathrm{SAB}$, then $\displaystyle \lim _{\mathrm{t} \rightarrow 1} \mathrm{k}$ is equal to :
Official answer
(4)
$\displaystyle \frac{13}{18}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.