JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the line \(\mathrm{L}: \sqrt{2} \mathrm{x}+\mathrm{y}=\alpha\) pass through the point of the intersection \(\mathrm{P}\) (in the first quadrant) of the circle \(x^2+y^2=3\) and the parabola \(x^2=2 y\). Let the line \(L\) touch two circles \(C_1\) and \(C_2\) of equal radius \(2 \sqrt{3}\). If the centres \(Q_1\) and \(Q_2\) of the circles \(C_1\) and \(C_2\) lie on the \(y\)-axis, then the square of the area of the triangle \(\mathrm{PQ}_1 \mathrm{Q}_2\) is equal to ...........
- A \(70\)
- B \(72\)
- C \(77\)
- D \(75\)
Answer & Solution
Correct Answer
(B) \(72\)
Step-by-step Solution
Detailed explanation
\(x^2+y^2=3 \text { and } x^2=2 y\) \(y^2+2 y-3=0 \Rightarrow(y+3)(y-1)=0\) \(y=-3 \text { or } y=1\) \(\mathrm{y}=1 \mathrm{x}=\sqrt{2} \Rightarrow P(\sqrt{2}, 1)\) \(\mathrm{p}\) lies on the line \(\sqrt{2} x+y=\alpha\) \(\sqrt{2}(\sqrt{2})+1=\alpha\) \(\alpha=3\) For circle…
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