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JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Ellipse

The length of the chord of the ellipse (x^2)/(4)+(y^2)/(2)=1 ,...

The length of the chord of the ellipse

\frac{x^2}{4}+\frac{y^2}{2}=1

, whose mid-point is \left(1, \frac{1}{2}\right)$, is :

Options

A $\frac{5}{3} \sqrt{15}$
B $\frac{1}{3} \sqrt{15}$
C $\frac{2}{3} \sqrt{15}$
D $\sqrt{15}$

Correct answer

C. $\frac{2}{3} \sqrt{15}$

Step-by-step solution

\begin{aligned} & T=S_1 \\ & \frac{x \cdot 1}{4}+\frac{y}{4}=\frac{1}{4}+\frac{1}{8} \\ & \Rightarrow 2 x+2 y=3 \\ & \frac{x^2}{4}+\frac{\left(\frac{3-2 x}{2}\right)^2}{2}=1 \\ & \Rightarrow x=\frac{12 \pm \sqrt{120}}{12} \Rightarrow y=\frac{1}{2} \mp \frac{\sqrt{120}}{12} \end{aligned}

So length of chord

=\frac{2 \sqrt{15}}{3}

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