Q34 / 75 · JEE Main 2025 (23 Jan Shift 2) · Mathematics

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 :

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

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

Hence, the correct answer is option: 'C'.

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