JEE Main 2025 — Ellipse Question with Solution

Given that the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is 10, we have:

\frac{2b^2}{a} = 10 \quad \Rightarrow \quad 5a = b^2 \tag{1}

Next, consider the function $f(t)=t^2+t+\frac{11}{12}$. To find its minimum value, we calculate the derivative:

$\frac{df(t)}{dt}=2t+1=0\Rightarrow t=\frac{-1}{2}$

Plugging $t=-\frac{1}{2}$ into $f(t)$ gives the minimum value:

$f\left(-\frac{1}{2}\right)={\left(-\frac{1}{2}\right)}^2+\left(-\frac{1}{2}\right)+\frac{11}{12}=\frac{1}{4}-\frac{1}{2}+\frac{11}{12}=\frac{3-6+11}{12}=\frac{8}{12}=\frac{2}{3}$

Thus, the eccentricity $e$ of the ellipse is $\frac{2}{3}$, so $e^2={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$.

Using the eccentricity formula for an ellipse:

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

Rearranging gives:

$b^2=a^2\left(1-\frac{4}{9}\right)=a^2\cdot \frac{5}{9}$

From equation $(1)$, $b^2=5a$. Substituting, we have:

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

Solving for $a$,

$a=9,\text{and therefore }{b}^2=5a=45\Rightarrow b=\sqrt{45}=3\sqrt{5}$

Finally, calculate $a^2+b^2$:

$a^2+b^2=81+45=126$