Ellipse – JEE Main 24 January 2025 Shift 1

Solution:

Let $P(x_1, y_1) = (\sqrt{3}, \frac{1}{2})$. The focal distances of a point on the ellipse are $a \pm ex_1$.
The product of focal distances is given by:
$$ (a – ex_1)(a + ex_1) = a^2 – e^2x_1^2 $$
Given product = $\frac{7}{4}$, and $x_1 = \sqrt{3}$:
$$ a^2 – 3e^2 = \frac{7}{4} \Rightarrow a^2 = \frac{7}{4} + 3e^2 \quad \dots(1) $$

Also, the point $P(\sqrt{3}, \frac{1}{2})$ lies on the ellipse:
$$ \frac{3}{a^2} + \frac{1/4}{b^2} = 1 $$
We know $b^2 = a^2(1-e^2)$.
$$ \frac{3}{a^2} + \frac{1}{4a^2(1-e^2)} = 1 $$
Multiply by $4a^2(1-e^2)$:
$$ 12(1-e^2) + 1 = 4a^2(1-e^2) $$
$$ 13 – 12e^2 = 4a^2(1-e^2) $$

Substitute $a^2$ from (1):
$$ 13 – 12e^2 = 4(\frac{7}{4} + 3e^2)(1-e^2) $$
$$ 13 – 12e^2 = (7 + 12e^2)(1-e^2) $$
$$ 13 – 12e^2 = 7 – 7e^2 + 12e^2 – 12e^4 $$
$$ 13 – 12e^2 = 7 + 5e^2 – 12e^4 $$
$$ 12e^4 – 17e^2 + 6 = 0 $$

Solve quadratic in $e^2$:
$$ e^2 = \frac{17 \pm \sqrt{289 – 288}}{24} = \frac{17 \pm 1}{24} $$
Two possible values for $e^2$:
$$ e_1^2 = \frac{18}{24} = \frac{3}{4} \Rightarrow e_1 = \frac{\sqrt{3}}{2} $$
$$ e_2^2 = \frac{16}{24} = \frac{2}{3} \Rightarrow e_2 = \sqrt{\frac{2}{3}} $$

The absolute difference of eccentricities is:
$$ |e_1 – e_2| = \left| \frac{\sqrt{3}}{2} – \frac{\sqrt{2}}{\sqrt{3}} \right| $$
$$ = \frac{3 – 2\sqrt{2}}{2\sqrt{3}} $$

Ans. (3)