JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(C\) be the circle of minimum area enclosing the ellipse \(E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with eccentricity \(\frac{1}{2}\) and foci \(( \pm 2,0)\). Let PQR be a variable triangle, whose vertex \(P\) is on the circle \(C\) and the side \(Q R\) of length 29 is parallel to the major axis of \(E\) and contains the point of intersection of \(E\) with the negative \(y\)-axis. Then the maximum area of the triangle PQR is :
- A \(6(3+\sqrt{2})\)
- B \(8(3+\sqrt{2})\)
- C \(62+\sqrt{3}\)
- D \(82+\sqrt{3}\)
Answer & Solution
Correct Answer
(D) \(82+\sqrt{3}\)
Step-by-step Solution
Detailed explanation
Area of \(\triangle P Q R\) \(\begin{aligned} & =\frac{1}{2}(2 a)(a \sin \theta+b) \\ & \therefore \text { maximum area }=a(a+b) \\ & \quad=4(4+2 \sqrt{3})=8(2+\sqrt{3}) \end{aligned}\)
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