JEE Mathematics 2025 Question with Solution

Given: $P(4, 4\sqrt{3})$ lies on the parabola $y^2 = 4ax$, and $PQ$ is a focal chord. Points $M$ and $N$ are the feet of perpendiculars from $P$ and $Q$ on the directrix.

Find: The area of quadrilateral $PQMN$.

Since $P(4, 4\sqrt{3})$ lies on $y^2 = 4ax$,

So,

and the parabola is

For $y^2 = 4ax$ with $4a = 12$, we get

A point on the parabola in parametric form is $\left(at^2, 2at\right)$. For point $P(4, 4\sqrt{3})$,

Since $PQ$ is a focal chord, the parameters satisfy

Hence,

Therefore, the coordinates of $Q$ are

The directrix of $y^2 = 4ax$ is

Since perpendiculars are drawn to the directrix, $PM$ and $QN$ are horizontal distances from the points to $x = -3$. Thus,

Also, $MN$ is the distance between the projections of $P$ and $Q$ on the vertical line $x = -3$, so it equals the difference of their $y$-coordinates:

Now $PQMN$ is a trapezium with parallel sides $PM$ and $QN$, and height $MN$. Therefore,

Therefore, the area of quadrilateral $PQMN$ is $\frac{343\sqrt{3}}{8}$. The correct option is B.

Note: The answer key marks option A, but the solution concludes option A while computing $\frac{343\sqrt{3}}{8}$, which matches option B. Using the solution working, the defensible answer is B.

Given: The parabola is $y^2 = 12x$, with directrix $x = -3$. The focal chord has endpoints $P(4, 4\sqrt{3})$ and $Q\left(\frac{9}{4}, -3\sqrt{3}\right)$.

Find: Area of $PQMN$.

Because the directrix is a vertical line, the perpendiculars from $P$ and $Q$ to the directrix are horizontal. Hence $PM$ and $QN$ are parallel, and their lengths are the horizontal distances of $P$ and $Q$ from the directrix.

So,

The segment joining their feet on the directrix is vertical, with length

Thus the quadrilateral is a trapezium with parallel sides $PM$ and $QN$ and distance between them equal to $MN$. Therefore,

Hence the correct option is B.