WBJEE · Maths · Ellipse
Let the foci of the ellipse \(\frac{x^{2}}{9}+y^{2}=1\) subtend a right angle at a point \(P\). Then, the locus of \(P\) is
- A \(x^{2}+y^{2}=1\)
- B \(x^{2}+y^{2}=2\)
- C \(x^{2}+y^{2}=4\)
- D \(x^{2}+y^{2}=8\)
Answer & Solution
Correct Answer
(D) \(x^{2}+y^{2}=8\)
Step-by-step Solution
Detailed explanation
\(\frac{x^{2}}{9}+\frac{y^{2}}{1}=1 \Rightarrow e=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}\) 0) ie, \((\pm 2 \sqrt{2}, 0)\) Two foci are \((\pm a e,\) Let \(P(h, k)\) by any point on the ellipse \(\therefore\) \(\frac{k-0}{h-2 \sqrt{2}} \times \frac{k-0}{h+2 \sqrt{2}}=-1\)…
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