Mathematics · 2022
JEE Main · 25 July 2022, Shift 2 · Q14
Let the foci of the ellipse x^2/16+y^2/7=1 and the hyperbola x^2/144-y^2/(α)=1/25 coincide. Then the length of the latus rectum of the hyperbola is:
Let the foci of the ellipse $\displaystyle \frac{x^2}{16}+\frac{y^2}{7}=1$ and the hyperbola $\displaystyle \frac{x^2}{144}-\frac{y^2}{\alpha}=\frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is :
Official answer
(4)
$\displaystyle \frac{27}{10}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.