WBJEE · Maths · Ellipse
The line \(x=2 y\) intersects the ellipse \(\frac{x^{2}}{4}+y^{2}=1\) at the points \(P\) and \(Q .\) The equation of the circle with \(P Q\) as diameter is
- A \(x^{2}+y^{2}=\frac{1}{2}\)
- B \(x^{2}+y^{2}=1\)
- C \(x^{2}+y^{2}=2\)
- D \(x^{2}+y^{2}=\frac{5}{2}\)
Answer & Solution
Correct Answer
(D) \(x^{2}+y^{2}=\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
Solving \(x=2 y\) and \(\frac{x^{2}}{4}+y^{2}=1\) Put \(x=2 y\) in Eq. (ii), we get \(\frac{(2 y)^{2}}{4}+y^{2}=1 \Rightarrow \frac{4 y^{2}}{4}+y^{2}=1\) \(=\) \(2 y^{2}=1 \Rightarrow y=\pm \frac{1}{\sqrt{2}}\) \(\therefore\) From \(\mathrm{E} q\) (i), \(x=\pm \sqrt{2}\)…
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