Let C be the circle of minimum area enclosing the ellipse with eccentricity and foci . Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 2a is parallel to the major axis of E and contains the point of intersecti

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Solution

Q.
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 QR of length 2a 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 : [2025]

1 $6 (2 + \sqrt{3})$

2 $8 (3 + \sqrt{2})$

3 $6 (3 + \sqrt{2})$

4 $8 (2 + \sqrt{3})$

Ans.
(4)

Given, foci $= (\pm 2, 0) = (\pm a e, 0)$ and eccentricity $(e) = \frac{1}{2}$

$∴ a e = 2$

$\Rightarrow a \times \frac{1}{2} = 2 \Rightarrow a = 4$ $[∵ e = \frac{1}{2}]$

Now, $b^{2} = a^{2} (1 - e^{2})$

$= 16 (1 - \frac{1}{4}) = 12 \Rightarrow b = 2 \sqrt{3}$

Since, the circle of minimum area enclosing the ellipse has a radius equal to semi-major axis i.e., a

$∴$ Radius = a = 4

Height of $△ P Q R = radius + 2 \sqrt{3} = 4 + 2 \sqrt{3}$

Hence, required area

$= \frac{1}{2} \times base \times height = \frac{1}{2} \times 2 a \times (4 + 2 \sqrt{3})$

$= 4 (4 + 2 \sqrt{3}) = 8 (2 + \sqrt{3}) sq . units$ .

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