Mathematics · 2024
JEE Main · 27 January 2024, Shift 2 · Q14
Let e_1 be the eccentricity of the hyperbola x^2/16-y^2/9=1 and e_2 be the eccentricity of the ellipse x^2/(a^2)+y^2/(b^2)=1, a > b, which passes…
Let $\displaystyle \mathrm{e}_1$ be the eccentricity of the hyperbola $\displaystyle \frac{x^2}{16}-\frac{y^2}{9}=1$ and $\displaystyle \mathrm{e}_2$ be the eccentricity of the ellipse $\displaystyle \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$, which passes through the foci of the hyperbola. If $\displaystyle \mathrm{e}_1 \mathrm{e}_2=1$, then the length of the chord of the ellipse parallel to the $\displaystyle x$-axis and passing through $\displaystyle (0,2)$ is :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \frac{10 \sqrt{5}}{3}$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.