JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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 \(\frac{1}{3} \sqrt{15}\)
- C \(\frac{2}{3} \sqrt{15}\)
- D \(\sqrt{15}\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{3} \sqrt{15}\)
Step-by-step Solution
Detailed explanation
\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}…
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