Maths Q12 – JEE-MAIN 2025

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

e = \frac{1}{2}

and foci

(\pm 2, 0)

. Let

P Q R

be a variable triangle, whose vertex

P

is on the circle

C

and the side

Q R

of length

2 a

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 Q R

is :

Clue 1 of 3: Understand Ellipse Properties

First, determine the parameters of the ellipse (a, b, c, e) from the given information about its eccentricity and foci.

🧠 Full Solution Path

Step 1: Determine Ellipse Parameters and Circle C✦ Active

Given the eccentricity $e = \frac{1}{2}$ and foci $(\pm c, 0) = (\pm 2, 0)$ , we have $c = 2$ . For an ellipse, $c = a e$ . Substituting the values, $2 = a \cdot \frac{1}{2}$ , which gives $a = 4$ . Now, find $b^{2}$ using $b^{2} = a^{2} (1 - e^{2})$ : $b^{2} = 4^{2} (1 - (\frac{1}{2})^{2}) = 16 (1 - \frac{1}{4}) = 16 \cdot \frac{3}{4} = 12$ . The ellipse equation is $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ . Since $a = 4$ and $b = \sqrt{12} = 2 \sqrt{3}$ , we have $a > b$ . The circle C of minimum area enclosing the ellipse is its auxiliary circle, $x^{2} + y^{2} = a^{2}$ . Thus, circle C is $x^{2} + y^{2} = 4^{2} = 16$ . The radius of C is $R = 4$ and its center is $(0, 0)$ .

Step 2: Determine Base QR of Triangle PQR○ Expand

The side QR has length $2 a = 2 (4) = 8$ . It is parallel to the major axis (x-axis) and contains the point of intersection of the ellipse E with the negative y-axis. To find this intersection, set $x = 0$ in the ellipse equation: $\frac{0^{2}}{16} + \frac{y^{2}}{12} = 1 ⟹ y^{2} = 12 ⟹ y = \pm 2 \sqrt{3}$ . The point on the negative y-axis is $(0, - 2 \sqrt{3})$ . Since QR is parallel to the x-axis and passes through $(0, - 2 \sqrt{3})$ , the line containing QR is $y = - 2 \sqrt{3}$ . Given that QR has length 8 and is centered on the y-axis, the coordinates of Q and R are $(- 4, - 2 \sqrt{3})$ and $(4, - 2 \sqrt{3})$ respectively. The base length of the triangle is $Q R = 8$ .

Step 3: Calculate Maximum Area of Triangle PQR○ Expand

Let $P (x_{P}, y_{P})$ be the vertex on circle C, so $x_{P}^{2} + y_{P}^{2} = 16$ . The height $h$ of the triangle PQR is the perpendicular distance from $P (x_{P}, y_{P})$ to the line $y = - 2 \sqrt{3}$ . So, $h = | y_{P} - (- 2 \sqrt{3}) | = | y_{P} + 2 \sqrt{3} |$ . The area of triangle PQR is $A = \frac{1}{2} \times base \times height = \frac{1}{2} \times 8 \times | y_{P} + 2 \sqrt{3} | = 4 | y_{P} + 2 \sqrt{3} |$ . To maximize the area, we need to maximize $| y_{P} + 2 \sqrt{3} |$ . Since $P$ is on the circle $x^{2} + y^{2} = 16$ , the range of $y_{P}$ is $[- 4, 4]$ . We need to find the maximum value of $| y_{P} + 2 \sqrt{3} |$ for $y_{P} \in [- 4, 4]$ . The expression $y_{P} + 2 \sqrt{3}$ ranges from $- 4 + 2 \sqrt{3}$ to $4 + 2 \sqrt{3}$ . Numerically, $2 \sqrt{3} \approx 3.464$ . So, the range is approximately $[- 0.536, 7.464]$ . The maximum absolute value will be $4 + 2 \sqrt{3}$ (when $y_{P} = 4$ ). Therefore, the maximum height $h_{m a x} = 4 + 2 \sqrt{3}$ . The maximum area is $A_{m a x} = 4 (4 + 2 \sqrt{3}) = 16 + 8 \sqrt{3} = 8 (2 + \sqrt{3})$ .

💡 Teacher's Secret Hint

Remember to consider the absolute value when calculating the height, as $y_{P} + 2 \sqrt{3}$ can be negative for some values of $y_{P}$ .

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Maths Question 12 – JEE-MAIN 2025

Clue 1 of 3: Understand Ellipse Properties

🧠 Full Solution Path