Let P(4, 4√3) Be a Point on the Parabola y² = 4ax and PQ Be a Focal Chord of the Parabola. If M and N Are the Feet of Perpendiculars Drawn From P and Q on the Directrix, Then the Area of the Quadrilateral PQMN Is Equal To

JEE Main 2025 — 22 January, Evening Shift. Previous Year Question.

Problem

Let $P(4,4\sqrt3)$ be a point on the parabola $y^2=4ax$ and $PQ$ be a focal chord of the parabola. If $M$ and $N$ are the feet of perpendiculars drawn from $P$ and $Q$ respectively on the directrix of the parabola, then the area of the quadrilateral $PQMN$ is equal to:

Key insight. $PM$ and $QN$ (the segments from $P$ and $Q$ to the directrix) are both horizontal, so $PQMN$ is actually a trapezoid — its area is just half the sum of these two parallel sides times the distance between them, no need for a general quadrilateral formula.

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Approach

Substitute $P(4,4\sqrt3)$ into $y^2=4ax$ to find $a$. Use the parametric form $(at^2,2at)$ to find $P$’s parameter, then use the standard focal-chord property that the two endpoints’ parameters multiply to $-1$ to find $Q$. Use the focal-distance property (distance from a point on the parabola to the directrix equals $x$-coordinate plus $a$) to get the trapezoid’s two parallel sides, and the vertical separation of $M,N$ on the directrix for its height.

Solution

Step 1 — Find a by substituting P into the parabola’s equation

$$(4\sqrt3)^2 = 4a(4) \implies 48=16a \implies a=3$$

So the parabola is $y^2=12x$, with focus at $(3,0)$ and directrix $x=-3$.

Step 2 — Find P’s parameter using (3t², 6t)

$$3t^2=4 \implies t^2=\frac43, \qquad 6t=4\sqrt3 \implies t=\frac{2\sqrt3}{3}=\frac{2}{\sqrt3}$$

(both give the consistent value $t_P=\tfrac{2}{\sqrt3}$)

Step 3 — Find Q’s parameter using the focal-chord property

For a focal chord, the two endpoints’ parameters satisfy $t_P\cdot t_Q=-1$:

$$t_Q = -\frac{1}{t_P} = -\frac{\sqrt3}{2}$$

Step 4 — Find Q’s coordinates

$$Q = \left(3t_Q^2,\ 6t_Q\right) = \left(3\cdot\frac34,\ 6\cdot\left(-\frac{\sqrt3}2\right)\right) = \left(\frac94,\ -3\sqrt3\right)$$

Step 5 — Find PM and QN using the focal-distance property

Distance from a point $(x,y)$ on the parabola to the directrix $x=-a$ equals $x+a$:

$$PM = 4+3 = 7, \qquad QN = \frac94+3 = \frac{21}{4}$$

Step 6 — Find MN (the vertical separation on the directrix)

$M$ and $N$ share the same $y$-coordinates as $P$ and $Q$ respectively (since they’re horizontal feet of perpendicular onto the vertical directrix):

$$MN = \left|4\sqrt3-(-3\sqrt3)\right| = 7\sqrt3$$

Step 7 — Apply the trapezoid area formula

$PM$ and $QN$ are the two parallel sides (both horizontal), and $MN$ is the perpendicular distance between them:

$$\text{Area} = \frac12(PM+QN)\times MN = \frac12\left(7+\frac{21}{4}\right)\times7\sqrt3$$

Step 8 — Simplify

$$7+\frac{21}{4} = \frac{28+21}{4} = \frac{49}{4}$$

$$\text{Area} = \frac12\times\frac{49}{4}\times7\sqrt3 = \frac{49\times7\sqrt3}{8} = \frac{343\sqrt3}{8}$$

Answer

Common mistakes

• Using the general shoelace formula for a quadrilateral instead of recognising the trapezoid shape. Since $M$ and $N$ both lie on the vertical directrix and $PM$, $QN$ are both horizontal, $PQMN$ is guaranteed to be a trapezoid — the simpler trapezoid formula avoids unnecessary coordinate-geometry bookkeeping.
• Getting the focal-chord parameter relation backwards. The standard property is $t_1t_2=-1$ for the two ends of a focal chord (not $t_1=t_2$ or $t_1+t_2=-1$) — using the wrong relation gives a completely different point for $Q$.

Practise next

• Let $P(9,6\sqrt2)$ be a point on the parabola $y^2=4ax$ and $PQ$ be a focal chord. Find the area of the quadrilateral $PQMN$ formed with the feet of perpendiculars from $P,Q$ to the directrix, using the same approach.