JEE Mathematics 2025 Question with Solution

Given:

• $E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a>b$
• $E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ with $$A

$$$$

Given: Same data as above.

Find: The intended correct option from the extracted page.

The extracted solution contains multiple inconsistencies:

1. It states the product of latus rectums as $\sqrt{\frac{32}{3}}$ in the working, whereas the question states $\frac{32}{\sqrt{3}}$.
2. One part of the solution derives unrealistic relations such as $ab=1$ and an incorrect value of $b$.
3. Another part correctly uses

$$a^2-b^2=4, \qquad b^2=\frac{2}{3}a^2$$

which gives

$$a^2=12, \qquad b^2=8$$

4. Despite the final written numerical conclusion in the working being $12\sqrt{6}$, the page explicitly labels Option C as correct.

Hence, by the extraction rule that the solution is the primary source and its explicit option label is used when present, the accepted answer is C.

Therefore, the area of quadrilateral $ABCD$ is taken to be $\frac{12\sqrt{6}}{5}$.