The length of the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$, whose mid-point is $\left(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 :
  1. $\frac{5}{3} \sqrt{15}$
  2. $\frac{1}{3} \sqrt{15}$
  3. $\frac{2}{3} \sqrt{15}$
  4. $\sqrt{15}$

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

Asked in: JEE Main 2025 (23 Jan Shift 2)

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