Question

Let $A, B,$ and $C$ be three points on the parabola $y^2 = 6x$, and let the line segment $AB$ meet the line $L$ through $C$ parallel to the $x$-axis at the point $D$. Let $M$ and $N$ respectively be the feet of the perpendiculars from $A$ and $B$ on $L$. Then $\left( \frac{\text{AM} \cdot \text{BN}}{\text{CD}} \right)^2$ is equal to ______ .

Answer 1

We are given:
$m_{AB} = m_{AD}$
$\implies \frac{2}{t_1 + t_2} = \frac{2a(t_1 - t_3)}{a t_1^2 - \alpha}$
$\implies a t_1^2 - \alpha = a \\{ t_1^2 - t_1 t_3 + t_1 t_2 - t_2 t_3 \\}$
$\implies \alpha = a ( t_1 t_3 + t_2 t_3 - t_1 t_2 )$
$AM = |2a ( t_1 - t_3 )|, \quad BN = |2a ( t_2 - t_3 )|,$
$CD = |a t_3^2 - \alpha|$
$CD = |a t_3^2 - a ( t_1 t_3 + t_2 t_3 - t_1 t_2 ) |$
$= a | t_3 ( t_3 - t_1 ) - t_2 ( t_3 - t_1 ) |$
$= a | ( t_3 - t_2 ) ( t_3 - t_1 ) |$
$\left( \frac{AM \cdot BN}{CD} \right)^2 = \frac{ \\{ 2a ( t_1 - t_3 ) \cdot 2a ( t_2 - t_3 ) \\}^2 }{ a ( t_3 - t_2 ) ( t_3 - t_1 )}$
$= \frac{16 a^2}{a} \cdot \frac{( t_1 - t_3 )^2 ( t_2 - t_3 )^2 }{ ( t_3 - t_2 ) ( t_3 - t_1 )}$
$16 a^2 = 16 \times \frac{9}{4} = 36$

Thus, the final answer is:
$\boxed{36}$