JEE Mathematics 2025 Question with Solution

Given: $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and $H: \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ with $a-A=2$. The ratio of eccentricities is taken from the question as $\frac{e_E}{e_H}=\frac{1}{3}$.

Find: The sum of the lengths of the latus rectums of the ellipse and the hyperbola.

For the ellipse,

$$e_E = \frac{c_1}{a}, \quad c_1^2 = a^2-b^2$$

and for the hyperbola,

$$e_H = \frac{c_2}{A}, \quad c_2^2 = A^2+B^2.$$

the solution concludes with the latus rectum formulas

$$L_1 = \frac{2b^2}{a}, \quad L_2 = \frac{2B^2}{A}.$$

Using

$$b^2 = a^2(1-e_E^2), \quad B^2 = A^2(e_H^2-1),$$

one gets

$$L_1 = 2a(1-e_E^2), \quad L_2 = 2A(e_H^2-1).$$

The extracted working on the page contains inconsistencies: one approach states Option A but computes the sum as $8$, while the second approach also concludes $\boxed{8}$. Since the numerical working in the solution repeatedly concludes $8$, the defensible answer from the solution is option C.

Therefore, the sum of the lengths of the latus rectums is $8$, so the correct option is C.

Given: The conics are an ellipse and a hyperbola. The solution gives the standard relations for eccentricity and latus rectum.

Find: Which option matches the result supported by the actual working shown.

From the extracted solution text:

1. For the ellipse,

$$e_1^2 = 1 - \frac{b^2}{a^2}.$$

2. For the hyperbola,

$$e_2^2 = 1 + \frac{B^2}{A^2}.$$

3. Their latus rectum lengths are

$$L_1 = \frac{2b^2}{a}, \quad L_2 = \frac{2B^2}{A}.$$

Hence,

$$L_1 = 2a(1-e_1^2), \quad L_2 = 2A(e_2^2-1).$$

The first displayed approach on the page finally writes

$$\text{Sum of LR} = \frac{2b^2}{a} + \frac{2B^2}{A} = 8.$$

The second approach also ends with

$$\boxed{8}.$$

So although the solution says The Correct Option is A, the actual algebraic conclusion shown in both approaches is $8$. Since option C is $8$, that is the answer supported by the solution content.

Therefore, the correct option is C.