For a parabola y²=4ax, any point can be represented parametrically as (at², 2at). We found a=3, so P(3t_1², 6t_1). Using the coordinates of P, we find t_1 = (2√3)/(3). Since PQ is a focal chord, the product of the parameters t_1 and t_2 is -1. This allows us to find t_2 = -(√3)/(2). Substituting t_2 back into the parametric form gives the coordinates of Q as ((9)/(4), -3√3).

For a parabola y²=4ax, any point can be represented parametrically as (at², 2at). We found a=3, so P(3t_1², 6t_1). Using the coordinates of P, we find t_1 = (2√3)/(3). Since PQ is a focal chord, the product of the parameters t_1 and t_2 is -1. This allows us to find t_2 = -(√3)/(2). Substituting t_2 back into the parametric form gives the coordinates of Q as ((9)/(4), -3√3).

The quadrilateral PQMN is a trapezium because PM and QN are perpendicular to the directrix, making them parallel to each other. The area of a trapezium is given by (1)/(2) (sum of parallel sides) × (height). Here, PM and QN are the parallel sides, and MN is the height. We calculate the lengths of PM, QN, and MN and substitute them into the formula to find the area.

The quadrilateral PQMN is a trapezium because PM and QN are perpendicular to the directrix, making them parallel to each other. The area of a trapezium is given by (1)/(2) (sum of parallel sides) × (height). Here, PM and QN are the parallel sides, and MN is the height. We calculate the lengths of PM, QN, and MN and substitute them into the formula to find the area.

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 foot of perpendiculars drawn from $P$ and $Q$ respectively on the directrix of the parabola, then the area of the quadrilateral $PQMN$ is equal to:

Answer: The area of the quadrilateral $PQMN$ is $\frac{343\sqrt{3}}{8}$ square units.

Step-by-step solution

Step 1: Determine the parabola's parameter 'a' and coordinates of P

The given point $P(4{,}4\sqrt{3})$ lies on the parabola $y^2=4ax$ . We can substitute the coordinates of $P$ into the parabola's equation to find the value of the parameter $a$ . This gives us $a=3$ . Thus, the equation of the parabola is $y^2=12x$ . The focus is $F(a,0) = F(3{,}0)$ and the directrix is $x=-a$ , which is $x=-3$ .

Step 2: Find the coordinates of point Q using focal chord property

For a parabola $y^2=4ax$ , any point can be represented parametrically as $(at^2, 2at)$ . We found $a=3$ , so $P(3t_1^2, 6t_1)$ . Using the coordinates of $P$ , we find $t_1 = \frac{2\sqrt{3}}{3}$ . Since $PQ$ is a focal chord, the product of the parameters $t_1$ and $t_2$ is $-1$ . This allows us to find $t_2 = -\frac{\sqrt{3}}{2}$ . Substituting $t_2$ back into the parametric form gives the coordinates of $Q$ as $(\frac{9}{4}, -3\sqrt{3})$ .

Step 3: Determine the coordinates of M and N on the directrix

The directrix of the parabola is the line $x=-3$ . Since $M$ and $N$ are the feet of the perpendiculars from $P$ and $Q$ to the directrix, their $x$ -coordinates will be $-3$ , and their $y$ -coordinates will be the same as $P$ and $Q$ respectively. So, $M(-3, 4\sqrt{3})$ and $N(-3, -3\sqrt{3})$ .

Step 4: Calculate the area of the quadrilateral PQMN

The quadrilateral $PQMN$ is a trapezium because $PM$ and $QN$ are perpendicular to the directrix, making them parallel to each other. The area of a trapezium is given by $\frac{1}{2} (\text{sum of parallel sides}) \times (\text{height})$ . Here, $PM$ and $QN$ are the parallel sides, and $MN$ is the height. We calculate the lengths of $PM$ , $QN$ , and $MN$ and substitute them into the formula to find the area.

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