JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let an ellipse \(E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}\), passes through \(\left(\sqrt{\frac{3}{2}}, 1\right)\) and has ecentricity \(\frac{1}{\sqrt{3}} .\) If a circle, centered at focus \(\mathrm{F}(\alpha, 0), \alpha>0\), of \(\mathrm{E}\) and radius \(\frac{2}{\sqrt{3}}\), intersects \(\mathrm{E}\) at two points \(\mathrm{P}\) and \(\mathrm{Q}\), then \(\mathrm{PQ}^{2}\) is equal to:
- A \(\frac{8}{3}\)
- B \(\frac{4}{3}\)
- C \(3\)
- D \(\frac{16}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{16}{3}\)
Step-by-step Solution
Detailed explanation
\(\frac{3}{2 a^{2}}+\frac{1}{b^{2}}=1 \text { and } 1-\frac{b^{2}}{a^{2}}=\frac{1}{3}\) \(\Rightarrow a^{2}=3 b^{2}=2\) \(\Rightarrow \frac{x^{2}}{3}+\frac{y^{2}}{2}=1.....(i)\) Its focus is \((1,0)\) Now, eqn of circle is \((x-1)^{2}+y^{2}=\frac{4}{3}.....(ii)\) Solving \((i)\)…
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