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

Correct Answer: 36

\[ m_{AB} = m_{AD} \] \[ \Rightarrow \frac{2}{t_1 + t_2} = \frac{2a(t_1 - t_3)}{a t_1^2 - \alpha} \] \[ \Rightarrow a t_1^2 - \alpha = a \{ t_1^2 - t_1 t_3 + t_1 t_2 - t_2 t_3 \} \] \[ \Rightarrow \alpha = a (t_1 t_3 + t_2 t_3 - t_1 t_2) \] \[ AM = |2a (t_1 - t_3)|, \quad BN = |2a (t_2 - t_3)| \] \[ CD = |a t_3^2 - \alpha| \] \[ CD = |a t_3^2 - a (t_1 t_3 + t_2 t_3 - t_1 t_2)| \] \[ = a |t_3^2 - t_1 t_3 - t_2 t_3 + t_1 t_2| \] \[ = a |t_3 (t_3 - t_1 - t_2) + t_1 t_2| \] \[ CD = a |(t_3 - t_2)(t_3 - t_1)| \] \[ \left( \frac{AM \cdot BN}{CD} \right)^2 = \left\{ \frac{2a(t_1 - t_3) \cdot 2a(t_2 - t_3)}{a(t_3 - t_2)(t_3 - t_1)} \right\}^2 \] \[ 16a^2 = 16 \times \frac{9}{4} = 36 \] 

Answer 2

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:
\[ \boxed{36} \]