Let $P(4, 4\sqrt{3})$ 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:
- (1) $\frac{263\sqrt{3}}{8}$
- (2) $17\sqrt{3}$
- (3) $\frac{343\sqrt{3}}{8}$
- (4) $\frac{34\sqrt{3}}{3}$
Solution:
Point $P(4, 4\sqrt{3})$ lies on $y^2 = 4ax$:
$$ (4\sqrt{3})^2 = 4a(4) \Rightarrow 48 = 16a \Rightarrow a = 3 $$
Equation is $y^2 = 12x$. Directrix is $x = -3$.
Parametric coordinates $(at^2, 2at)$. For $P$:
$$ 2(3)t_1 = 4\sqrt{3} \Rightarrow 6t_1 = 4\sqrt{3} \Rightarrow t_1 = \frac{2}{\sqrt{3}} $$
Since $PQ$ is a focal chord, $t_1t_2 = -1 \Rightarrow t_2 = -\frac{\sqrt{3}}{2}$.
Coordinates of $Q(at_2^2, 2at_2)$:
$$ x_Q = 3\left(-\frac{\sqrt{3}}{2}\right)^2 = 3\left(\frac{3}{4}\right) = \frac{9}{4} $$
$$ y_Q = 6\left(-\frac{\sqrt{3}}{2}\right) = -3\sqrt{3} $$
Quadrilateral $PQMN$ is a trapezium with parallel sides $PM$ and $QN$ (perpendicular to directrix) and height $MN$ (segment on directrix).
Lengths of parallel sides (distance from $x = -3$):
$$ PM = 4 – (-3) = 7 $$
$$ QN = \frac{9}{4} – (-3) = \frac{21}{4} $$
Height $MN$ is the vertical distance between $P$ and $Q$:
$$ MN = 4\sqrt{3} – (-3\sqrt{3}) = 7\sqrt{3} $$
Area of Trapezium:
$$ \text{Area} = \frac{1}{2}(PM + QN) \times MN $$
$$ = \frac{1}{2}\left(7 + \frac{21}{4}\right) \times 7\sqrt{3} = \frac{1}{2}\left(\frac{49}{4}\right) \times 7\sqrt{3} = \frac{343\sqrt{3}}{8} $$
Ans. (3)