JEE Main 2025 Jan 24 Shift 1, Mathematics Q10: Introduction to Ellipse

For an ellipse, the product of focal distances at the point $(x_1,y_1)$ is $a^2-e^2{x}_1^2$. Here $x_1=\sqrt{3}$, so

$a^2-3e^2=\frac{7}{4}\Rightarrow 4a^2=7+12e^2(1)$

The point also lies on the ellipse:

$\frac{3}{a^2}+\frac{1}{4b^2}=1$, with $b^2=a^2(1-e^2)$:

$\frac{3}{a^2}+\frac{1}{4a^2(1-e^2)}=1$

Multiply through by $4a^2(1-e^2)$ and use $(1)$:

$12(1-e^2)+1=4a^2(1-e^2)=(7+12e^2)(1-e^2)$

Expand and simplify:

$13-12e^2=7-7e^2+12e^2-12e^4$

$12e^4-17e^2+6=0$

Quadratic in $e^2$:

$e^2=\frac{17\pm \sqrt{289-288}}{24}=\frac{17\pm 1}{24}=\frac{3}{4}\text{or}\frac{2}{3}$

So $e_1=\frac{\sqrt{3}}{2}$ and $e_2=\sqrt{\frac{2}{3}}$.

$|e_1-e_2|=\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2\sqrt{2}}{2\sqrt{3}}$ (after rationalising the second term and combining over $2\sqrt{3}$).

$\therefore$ Option (3).