Question

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

• A. \( \frac{343\sqrt{3}}{8} \)
• B. \( \frac{343\sqrt{3}}{8} \)
• C. \( \frac{34\sqrt{3}}{3} \)
• D. \( 17\sqrt{3} \)

Hint:

For problems involving parabolas and focal chords: - Use the standard equation of the parabola and the properties of the focal chord (e.g., the relationship between the point on the parabola and the focus). - Employ geometric properties to find areas, especially when perpendiculars from points on the parabola are involved.

Answer 1

The Correct Option is A

Given that the point \( (4, 4\sqrt{3}) \) satisfies the equation \( y^2 = 4ax \), we substitute the coordinates: \( (4\sqrt{3})^2 = 4a(4) \). This simplifies to \( 48 = 16a \), which yields \( 4a = 12 \). Consequently, the equation of the parabola is \( y^2 = 12x \). The parameter for point \( P \) is given as \( t_1 = \frac{2}{\sqrt{3}} \). For point \( Q \), the parameter is \( t_2 = -\frac{\sqrt{3}}{2} \), resulting in the coordinates \( \left( \frac{9}{4}, -3\sqrt{3} \right) \). \textbf{The area of trapezium PQNM is calculated as follows:} \[ \text{Area} = \frac{1}{2} \times MN \times (PM + QN) \] This can be expressed as: \[ \text{Area} = \frac{1}{2} \times MN \times (PS + QS) \] Further simplification leads to: \[ \text{Area} = \frac{1}{2} \times MN \times PQ \] Substituting the known values: \[ \text{Area} = \frac{1}{2} \times 7\sqrt{3} \times \frac{49}{4} \] The final area is: \[ \text{Area} = \frac{343\sqrt{3}}{8} \]