JEE Mathematics 2025 Question with Solution

Given: The lines $y = x + p$ touch the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ at points $A$ and $B$. The line $y = x$ intersects the ellipse at points $C$ and $D$.

Find: The area of quadrilateral $ABCD$.

From the solution, the points of contact are

$$A\left(\frac{-16}{5}, \frac{9}{5}\right), \quad B\left(\frac{16}{5}, \frac{-9}{5}\right)$$

and one intersection point is

$$D\left(\frac{12}{5}, \frac{12}{5}\right).$$

The area of triangle $ABD$ is calculated by the determinant formula:

$$\text{Area of } ABD = \frac{1}{2} \left| \begin{array}{ccc} \frac{-16}{5} & \frac{9}{5} & 1 \\ \frac{16}{5} & \frac{-9}{5} & 1 \\ \frac{12}{5} & \frac{12}{5} & 1 \\ \end{array} \right| = 12$$

Since the quadrilateral is formed symmetrically, the area of quadrilateral $ABCD$ is twice the area of triangle $ABD$:

$$\text{Area of } ABCD = 2 \times 12 = 24$$

Therefore, the area of the quadrilateral is $24$, so the correct option is B.

Given: Coordinates stated in the solution are $A\left(\frac{-16}{5}, \frac{9}{5}\right)$, $B\left(\frac{16}{5}, \frac{-9}{5}\right)$ and $D\left(\frac{12}{5}, \frac{12}{5}\right)$.

Find: The area of quadrilateral $ABCD$.

Use the determinant form for the area of a triangle with vertices $\left(x_1,y_1\right)$, $\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$:

$$\text{Area} = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|$$

Substituting for triangle $ABD$,

$$\text{Area of } ABD = \frac{1}{2} \left| \begin{array}{ccc} \frac{-16}{5} & \frac{9}{5} & 1 \\ \frac{16}{5} & \frac{-9}{5} & 1 \\ \frac{12}{5} & \frac{12}{5} & 1 \end{array} \right| = 12$$

Then the quadrilateral area is

$$\text{Area of } ABCD = 2 \times \text{Area of } ABD = 24$$

Hence, the correct option is B.