Let the length of a latus rectum of an ellipse \fracx^2a^2 + \fracy^2b^2 = 1 be 10. If its eccentricity is the minimum value of the function f(t) = t^2 + t + \frac1112 , t \in \mathbbR , then a^2 + b^2 is equal to :

Ellipse PYQ — JEE Main 2025 · 7 Apr · Shift 2 · Q33 with Solution

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Correct answer: D

Given ellipse and latus rectum

For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \qquad (a>b)$ its eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}$ and the length of its latus rectum is $\frac{2b^2}{a}.$

We are given: $\frac{2b^2}{a}=10 \implies b^2=5a. \qquad (1)$

Find the minimum value of $f(t)$

Given $f(t)=t^2+t+\frac{11}{12}.$ Complete the square:

=\left(t+\frac12\right)^2+\frac{8}{12} =\left(t+\frac12\right)^2+\frac23.$$ Hence the minimum value is $$\min f(t)=\frac23.$$ The eccentricity is given to be this minimum value, so $$e=\frac23.$$ --- 3. **Use eccentricity relation** For an ellipse, $$b^2=a^2(1-e^2).$$ Since $$e=\frac23 \implies e^2=\frac49,$$ we get $$b^2=a^2\left(1-\frac49\right)=a^2\cdot \frac59= rac59 a^2. \qquad (2)$$ --- 4. **Solve for** $a$ and $b$ From (1) and (2): $$5a=\frac59 a^2.$$ Assuming $a\neq 0$, divide by $a$: $$5=\frac59 a \implies a=9.$$ Then $$b^2=5a=45,$$ and $$a^2=81.$$ Therefore, $$a^2+b^2=81+45=126.$$ --- 5. **Check options** The correct option is: $$\boxed{126}$$ which is **Option D**.

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