Question

Consider a parabola \( y^2 = 8x \). The directrix of parabola cuts x-axis at A and PQ is a focal chord of parabola. If slope of PA = 3/5 and abscissa of P is greater than 1, then the area of \(\Delta AQP\) is:

• A. 40
• B. 69/2
• C. 80/3
• D. 23

Hint:

For any focal chord $PQ$, the product of the parameters $t_1 t_2 = -1$. This is a standard property that simplifies finding the coordinates of the second point.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For the parabola \( y^2 = 8x \), we have \( 4a = 8 \implies a = 2 \).
The focus is \( S(2, 0) \) and the directrix is \( x = -2 \).
The directrix cuts the x-axis at \( A(-2, 0) \).
Step 2: Key Formula or Approach:
Let \( P(2t^2, 4t) \) be a point on the parabola in parametric form.
Since \( PQ \) is a focal chord, the coordinates of \( Q \) will be \( \left( \frac{2}{t^2}, -\frac{4}{t} \right) \) because for a focal chord the product of parameters satisfies \( t_P \cdot t_Q = -1 \).
Step 3: Detailed Explanation:
The slope of \( PA \) is given as \( \frac{3}{5} \):
\[ m_{PA} = \frac{4t - 0}{2t^2 - (-2)} = \frac{4t}{2t^2 + 2} = \frac{2t}{t^2 + 1} = \frac{3}{5} \]
\[ 10t = 3t^2 + 3 \implies 3t^2 - 10t + 3 = 0 \]
\[ (3t - 1)(t - 3) = 0 \implies t = 3 \text{ or } t = \frac{1}{3} \]
Given the abscissa of \( P \) (which is \( 2t^2 \)) is greater than 1:
If \( t = 3 \), \( 2(3^2) = 18>1 \) (Accepted).
If \( t = \frac{1}{3} \), \( 2(1/3)^2 = 2/9<1 \) (Rejected).
So, \( P = (18, 12) \).
As \( t = 3 \), point \( Q \) is \( \left( \frac{2}{3^2}, \frac{-4}{3} \right) = \left( \frac{2}{9}, -\frac{4}{3} \right) \).
Points are \( A(-2, 0) \), \( P(18, 12) \), and \( Q(2/9, -4/3) \).
Area of \( \Delta AQP = \frac{1}{2} |x_A(y_P - y_Q) + x_P(y_Q - y_A) + x_Q(y_A - y_P)| \)
\[ \text{Area} = \frac{1}{2} \left| -2\left(12 - \left(-\frac{4}{3}\right)\right) + 18\left(-\frac{4}{3} - 0\right) + \frac{2}{9}(0 - 12) \right| \]
\[ \text{Area} = \frac{1}{2} \left| -2\left(\frac{40}{3}\right) - 24 - \frac{24}{9} \right| = \frac{1}{2} \left| -\frac{80}{3} - 24 - \frac{8}{3} \right| \]
\[ \text{Area} = \frac{1}{2} \left| -\frac{88}{3} - \frac{72}{3} \right| = \frac{1}{2} \left| -\frac{160}{3} \right| = \frac{80}{3} \]
Step 4: Final Answer:
The area of the triangle is \( \frac{80}{3} \).