Mathematics · 2020
JEE Main · 4 September 2020, Shift 1 · Q65
Let x^2/(a^2)+y^2/(b^2)=1( a > b ) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function,…
Let $\displaystyle \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1(\mathrm{a}>\mathrm{b})$ be a given ellipse, length of whose latus rectum is $\displaystyle 10$ . If its eccentricity is the maximum value of the function, $\displaystyle \phi(\mathrm{t})=\frac{5}{12}+\mathrm{t}-\mathrm{t}^2$, then $\displaystyle \mathrm{a}^2+\mathrm{b}^2$ is equal to:
Official answer
(2)
$\displaystyle 126$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.