Question

Let \( P(\alpha, \beta) \) be a point on the parabola \( y^2 = 4x \). If \( P \) also lies on the chord of the parabola \( x^2 = 8y \) whose midpoint is \( \left( 1, \frac{5}{4} \right) \), then \( (\alpha - 28)(\beta - 8) \) is equal to ______.

Answer 1

Correct Answer: 192

Step 1. The equation of the parabola is \(x^2 = 8y\)

Step 2. The chord with midpoint \( (x_1, y_1) \) has the equation \( T = S_1 \):
  \(x x_1 - 4(y + y_1) = x_1^2 - 8y_1\)
   Substituting \( (x_1, y_1) = (1, \frac{5}{4}) \):  
  \(x - 4 \left( y + \frac{5}{4} \right) = 1 - 8 \cdot \frac{5}{4} = -9\)
  
  \(x - 4y = -4\)

Step 3. Since \( P(\alpha, \beta) \) lies on this chord and also on the parabola \( y = \frac{x^2}{4} \), we have:
  \(\alpha - 4\beta = -4\)
   \(\beta^2 = 4\alpha\)

Step 4. Solve equations (ii) and (iii):
  Substitute \( \alpha = \frac{\beta^2}{4} \) from (iii) into (ii):  
  \(\frac{\beta^2}{4} - 4\beta = -4\)
  \(\beta^2 - 16\beta + 16 = 0\)
 \((\beta - 8)^2 = 48\)
  \(\beta = 8 \pm 4\sqrt{3}\)
  \(\beta = 8 \pm 4\sqrt{3}\)

Step 5. Substitute \( \beta = 8 \pm 4\sqrt{3} \) back into equation (ii) to find \( \alpha \):
  For \(\beta = 8 + 4\sqrt{3}\):  
  For  \(\beta = 8 - 4\sqrt{3}\):
   \(\alpha = 4(8 - 4\sqrt{3}) - 4 = 28 - 16\sqrt{3}\)

Step 6. Therefore, the possible points \( ( \alpha, \beta ) \) are:
  \(( \alpha, \beta ) = (28 + 16\sqrt{3}, 8 + 4\sqrt{3}) \text{ and } (28 - 16\sqrt{3}, 8 - 4\sqrt{3})\)

Step 7. Calculate \( ( \alpha - 28 )( \beta - 8 ) \): 
 \(( \alpha - 28 )( \beta - 8 ) = (\pm 16\sqrt{3})(\pm 4\sqrt{3}) = 16 \cdot 4 \cdot 3 = 192\)
 Thus, \(( \alpha - 28 )( \beta - 8 ) = 192\).

The Correct Answer is: 192