Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse…

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(\mathrm{a} \gt \mathrm{b})$, be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is
  1. $\frac{1-\sqrt{3}}{\sqrt{2}}$
  2. $\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$
  3. $\frac{3-2 \sqrt{2}}{3 \sqrt{2}}$
  4. $\frac{1-2 \sqrt{2}}{\sqrt{3}}$

Solution

$\begin{aligned} & \text { Product of focal distances }=\left(a+e x_1\right)\left(a-e x_1\right) \\ & =a^2-e^2 x_1^2=a^2-e^2(3) \\ & =a^2-3 e^2=\frac{7}{4} \Rightarrow a^2=\frac{7}{4}+3 e^2 \\ & \Rightarrow 4 a^2=7+12 e^2 \\ & \&\left(\sqrt{3}, \frac{1}{2}\right) \text { lines on } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ & \therefore \frac{3}{a^2}+\frac{1}{4 b^2}=1\end{aligned}$
$\begin{aligned} & \frac{3}{\mathrm{a}^2}+\frac{1}{4\left(\mathrm{a}^2\right)\left(1-\mathrm{e}^2\right)}=1 \\ & 12\left(1-\mathrm{e}^2\right)+1=4 \mathrm{a}^2\left(1-\mathrm{e}^2\right) \\ & 13-12 \mathrm{e}^2=\left(7+12 \mathrm{e}^2\right)\left(1-\mathrm{e}^2\right) \\ & \Rightarrow 13-12 \mathrm{e}^2=7-7 \mathrm{e}^2+12 \mathrm{e}^2-12 \mathrm{e}^4 \\ & \Rightarrow 12 \mathrm{e}^4-17 \mathrm{e}^2+6=0 \\ & \therefore \mathrm{e}^2=\frac{17 \pm \sqrt{289-288}}{24}=\frac{17 \pm 1}{24}=\frac{3}{4} \& \frac{2}{3} \\ & \therefore \mathrm{e}=\frac{\sqrt{3}}{2} \& \sqrt{\frac{2}{3}} \\ & \therefore \text { difference }==\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2 \sqrt{2}}{2 \sqrt{3}}\end{aligned}$

Asked in: JEE Main 2025 (24 Jan Shift 1)

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