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

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