QJEEM25Ma605d: JEE Main 2025 Mathematics - Class 11 JEEM25Ma605d Solution: Let and . Let the distance between the foci of E and the foci…

We are given an ellipse

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

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

and

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

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

$3A-A=2 \quad\Longrightarrow\quad 2A=2 \quad\Longrightarrow\quad A=1.$

Thus,

$a=3.$

Next, for the ellipse we have

$a^2-b^2=3 \quad \Longrightarrow\quad 9-b^2=3 \quad\Longrightarrow\quad b^2=6.$

For the hyperbola,

$A^2+B^2=3 \quad \Longrightarrow\quad 1+B^2=3 \quad\Longrightarrow\quad B^2=2.$

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:

$L_E=\frac{2\times6}{3}=\frac{12}{3}=4,$

For the hyperbola:

$L_H=\frac{2\times2}{1}=4.$

The sum of the lengths of the latus rectums is then

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

Thus, the answer is

$8.$