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 \( y^2 = 8x \), \( 4a = 8 \implies a = 2 \). The focus is \( S(2, 0) \) and the directrix is \( x = -2 \). The point \( A \) is the intersection of the directrix and x-axis, so \( A = (-2, 0) \).
Step 2: Key Formula or Approach:
1. Let \( P = (2t^2, 4t) \). Slope of \( PA = \frac{4t - 0}{2t^2 - (-2)} = \frac{4t}{2t^2 + 2} = \frac{2t}{t^2 + 1} \).
2. Focal chord property: if \( P \) is \( (2t^2, 4t) \), then \( Q \) is \( \left(\frac{2}{t^2}, -\frac{4}{t}\right) \).
Step 3: Detailed Explanation:
1. Given slope of \( PA = 3/5 \):
\[ \frac{2t}{t^2 + 1} = \frac{3}{5} \implies 10t = 3t^2 + 3 \implies 3t^2 - 10t + 3 = 0 \] \[ (3t - 1)(t - 3) = 0 \implies t = 3 \text{ or } t = \frac{1}{3} \] 2. Abscissa of \( P \) (\( x_P = 2t^2 \)) is \( > 1 \). If \( t=3, x_P=18 \). If \( t=\frac{1}{3}, x_P=\frac{2}{9} \). Thus, \( t=3 \).
3. Points: \( A(-2, 0) \), \( P(18, 12) \), \( Q\left(\frac{2}{9}, -\frac{4}{3}\right) \).
4. Area of \( \Delta AQP \):
\[ \text{Area} = \frac{1}{2} \left| x_A(y_Q - y_P) + x_Q(y_P - y_A) + x_P(y_A - y_Q) \right| \] \[ = \frac{1}{2} \left| -2\left(-\frac{4}{3} - 12\right) + \frac{2}{9}(12 - 0) + 18\left(0 - \left(-\frac{4}{3}\right)\right) \right| \] \[ = \frac{1}{2} \left| -2\left(-\frac{40}{3}\right) + \frac{24}{9} + 24 \right| = \frac{1}{2} \left| \frac{80}{3} + \frac{8}{3} + 24 \right| = \frac{1}{2} \left| \frac{88}{3} + \frac{72}{3} \right| = \frac{1}{2} \cdot \frac{160}{3} = \frac{80}{3} \]
Step 4: Final Answer:
The area of \( \Delta AQP \) is \( \frac{80}{3} \).