JEE Main 2025 Apr 3 Shift 2, Mathematics Q17: Introduction to Circles

JEE Main2025Apr 3, Shift 2Mathematics

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 $P QR$ be a variable triangle, whose vertex $P$ is on the circle $C$ and the side $QR$ of length $8$ 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 $P QR$ is:

A
$6 (3 + 2)$
B
$8 (3 + 2)$
C
$6 (2 + 3)$
D
$8 (2 + 3)$

Solution

For the ellipse,

$e = \frac{c}{a} = \frac{1}{2}, c = 2$

$\Rightarrow a = 4, b^{2} = a^{2} - c^{2} = 16 - 4 = 12$

Hence, the minimum area circle enclosing the ellipse is

$x^{2} + y^{2} = 16$

The point where the ellipse meets the negative $y$ -axis is $(0, - 23)$ .

Since $QR$ is parallel to the major axis and has length $8$ , its equation is

$y = - 23$ .

The area of $△ P QR = \frac{1}{2} \times QR \times (perpendicular distance of P from QR)$ .

The perpendicular distance is maximum when $P$ is the highest point of the circle, i.e., $P = (0, 4)$ .

$∴ h = 4 - (- 23) = 4 + 23$

$A_{m a x} = \frac{1}{2} \times 8 \times (4 + 23)$

$= 8 (2 + 3)$

$8 (2 + 3)$

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q17: Introduction to Circles

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