Mathematics · 2024
JEE Main · 27 January 2024, Shift 1 · Q16
The length of the chord of the ellipse x^2/25+y^2/16=1, whose mid point is (1, 2/5), is equal to:
The length of the chord of the ellipse $\displaystyle \frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\displaystyle \left(1, \frac{2}{5}\right)$, is equal to:
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \frac{\sqrt{1691}}{5}$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.