JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(PQ\) be a focal chord of the parabola \(y^{2}=4 x\) such that it subtends an angle of \(\frac{\pi}{2}\) at the point \((3, 0)\). Let the line segment \(PQ\) be also a focal chord of the ellipse \(E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}\). If \(e\) is the eccentricity of the ellipse \(E\), then the value of \(\frac{1}{e^{2}}\) is equal to
- A \(1+\sqrt{2}\)
- B \(3+2 \sqrt{2}\)
- C \(1+2 \sqrt{3}\)
- D \(4+5 \sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(3+2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(PQ\) is focal chord \(m _{P R} \cdot m_{P Q}=-1\) \(\frac{2 t }{ t ^{2}-3} \times \frac{-2 / t }{\frac{1}{ t ^{2}}-3}=-1\) \(\left( t ^{2}-1\right)^{2}=0\) \(\Rightarrow t =1\) \(\Rightarrow P\) and \(Q\) must be end point of latus rectum: \(\quad P (1,2)\) and \(Q (1,-2)\)…
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