JEE Main 2025 — Ellipse Question with Solution

We are given an ellipse

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b,$$

and a hyperbola

$$\frac{x^2}{A^2}-\frac{y^2}{B^2}=1.$$

It is stated that “the distance between the foci of $E$ and the foci of $H$ is $2\sqrt{3}$.” A natural interpretation is that the foci of the ellipse are separated by

$$2\sqrt{a^2-b^2},$$

and those of the hyperbola by

$$2\sqrt{A^2+B^2},$$

and each distance is equal to $2\sqrt{3}$. That is, we have

$2\sqrt{a^2-b^2}=2\sqrt{3}⟹a^2-b^2=3,$

and

$2\sqrt{A^2+B^2}=2\sqrt{3}⟹A^2+B^2=3.$

We are also given that

$$a-A=2,$$

and that the ratio of the eccentricities is

$$\frac{e_E}{e_H}=\frac{1}{3},$$

where the eccentricity of the ellipse is

$$e_E=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{3}}{a},$$

and the eccentricity of the hyperbola is

$$e_H=\frac{\sqrt{A^2+B^2}}{A}=\frac{\sqrt{3}}{A}.$$

Thus the ratio becomes

$$\frac{e_E}{e_H}=\frac{\sqrt{3}/a}{\sqrt{3}/A}=\frac{A}{a}=\frac{1}{3}.$$

This implies

$$a=3A.$$

Now, using the condition $a-A=2$ together with $a=3A$, we get

$$3A-A=2⟹2A=2⟹A=1.$$

Thus,

$$a=3.$$

Next, for the ellipse we have

$$a^2-b^2=3⟹9-b^2=3⟹b^2=6.$$

For the hyperbola,

$$A^2+B^2=3⟹1+B^2=3⟹B^2=2.$$

The length of the latus rectum is given by the following formulas:

For the ellipse:

$$L_E=\frac{2b^2}{a},$$

For the hyperbola:

$$L_H=\frac{2B^2}{A}.$$

Substitute the computed values:

For the ellipse:

$$L_E=\frac{2\times 6}{3}=\frac{12}{3}=4,$$

For the hyperbola:

$$L_H=\frac{2\times 2}{1}=4.$$

The sum of the lengths of the latus rectums is then

$$L_E+L_H=4+4=8.$$

Thus, the answer is

$$8.$$