$\begin{aligned}
& T=S_1 \
& \frac{x \cdot 1}{4}+\frac{y}{4}=\frac{1}{4}+\frac{1}{8} \
& \Rightarrow 2 x+2 y=3 \
& \frac{x^2}{4}+\frac{\left(\frac{3-2 x}{2}\right)^2}{2}=1 \
& \Rightarrow x=\frac{12 \pm \sqrt{120}}{12} \Rightarrow y=\frac{1}{2} \mp \frac{\sqrt{120}}{12}
\end{aligned}Solengthofchord=\frac{2 \sqrt{15}}{3}$
The length of the chord of the ellipse x^2 4+ y^2 2=1, whose mid-point is (1, 1 2 ), is :
JEE Main 2025 — Mathematics Coordinate Geometry
The length of the chord of the ellipse 4x2+2y2=1, whose mid-point is (1,21), is :
Held on 23 Jan 2025 · Verified 6 Jul 2026.
Options
- A
3515
- B
3115
- C
3215
- D
15
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