JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The line \(y=x+1\) meets the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{2}=1\) at two points \(P\) and \(Q\). If \(r\) is the radius of the circle with \(PQ\) as diameter then \((3 r )^{2}\) is equal to
- A \(20\)
- B \(12\)
- C \(11\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(20\)
Step-by-step Solution
Detailed explanation
Ellipse \(x^{2}+2 y^{2}=4\) Line \(y=x+1\) Point of intersection \(x^{2}+2(x+1)^{2}=4\) \(3 x^{2}+4 x-2=0\) \(\left|x_{1}-x_{2}\right|=\frac{\sqrt{40}}{3}\) \(AB =2 r =\left| x _{1}- x _{2}\right| \sqrt{1+ m ^{2}}\) \(m\) is slope of given line…
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