JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the midpoint of a chord of the ellipse \(\frac{x^2}{9}+\frac{y^2}{4}=1\) is \((\sqrt{2}, 4 / 3)\), and the length of the chord is \(\frac{2 \sqrt{\alpha}}{3}\), then \(\alpha\) is :
- A \(20\)
- B \(22\)
- C \(18\)
- D \(26\)
Answer & Solution
Correct Answer
(B) \(22\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & E: \frac{x^2}{9}+\frac{y^2}{4}=1 \\ & T=S_1 \\ & \Rightarrow \frac{\sqrt{2} x}{9}+\frac{1}{4}\left(\frac{4}{3} y\right)-1=\frac{2}{9}+\frac{16}{9(4)}-1 \\ & \frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9} \\ & \frac{\sqrt{2}…
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