Mathematics · Ellipse

JEE Main 2025 — 7 April, Evening Shift — Question 35

Let the length of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ be 10 . If its eccentricity is the minimum value of

the function $f(t)=t^{2}+t+\frac{11}{12}, t \in \mathbb{R}$ , then $a^{2}+b^{2}$ is equal to

Answer: C

$\frac{2 b^{2}}{a}=10 \Rightarrow b^{2}=5 a$

$e=\left.f(t)\right|_{\text {min }}=t^{2}+t+\left.\frac{11}{12}\right|_{t=\frac{1}{4}}=\frac{1}{4}-\frac{1}{2}+\frac{11}{12}$

$=\frac{3-6+11}{12}=\frac{8}{12}=\frac{2}{3}$

$\Rightarrow \sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{2}{3}$

$\Rightarrow 1-\frac{b^{2}}{a^{2}}=\frac{4}{9} \Rightarrow \frac{b^{2}}{a^{2}}=\frac{5}{9} \Rightarrow b^{2}=\frac{5 a^{2}}{9}$

$\Rightarrow 5 a=\frac{5 a^{2}}{9} \Rightarrow a=9 ;$

So, $b^{2}=45$

Hence, $a^{2}+b^{2}=81+45=126$

Answer key and solution verified before publishing.

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