Let the product of the focal distances of the point ( 3 , (1)/(2)) on the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1,(a ) , be (7)/(4) . Then the absolute difference of the eccentricities of two such ellipses is

Correct answer: B. 3-2 2 2 3 aligned & Product of focal distances =(a+e x_1)(a-e x_1) & =a^2-e^2 x_1^2=a^2-e^2(3) & =a^2-3 e^2=(7)/(4) a^2=(7)/(4)+3 e^2 & 4 a^2=7+12 e^2 & &( 3 , (1)/(2)) lines on (x^2)/(a^2)+(y^2)/(b^2)=1 & (3)/(a^2)+(1)/(4 b^2)=1 aligned aligned & 3 a^2 + 1 4(a^2)(1-e^2) =1 & 12(1-e^2)+1=4 a^2(1-e^2) & 13-12 e^2=(7+12 e^2)(1-e^2) & 13-12 e^2=7-7 e^2+12 e^2-12 e^4 & 12 e^4-17 e^2+6=0 & e^2= 17 ± 289-288 24 =(17 ± 1)/(24)=(3)/(4) & (2)/(3) & e= 3 2 & (2)/(3) & difference == 3 2 - (2)/(3) = 3-2 2 2 3 aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Ellipse

Let the product of the focal distances of the point ( 3 , (1)/(2))...

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

Options

A $\frac{1-\sqrt{3}}{\sqrt{2}}$
B $\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$
C $\frac{3-2 \sqrt{2}}{3 \sqrt{2}}$
D $\frac{1-2 \sqrt{2}}{\sqrt{3}}$

Correct answer

B. $\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$

Step-by-step solution

\begin{aligned} & \text{ Product\text{ 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}

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