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 $$\left(\frac{AM \cdot BN}{CD}\right)^2$$ is equal to ________

Take points on parabola (y^2=6x) as
$$A(\frac{y_1^2}{6},,y_1),\quad B(\frac{y_2^2}{6},,y_2),\quad C(\frac{y_3^2}{6},,y_3)$$
Line (L) through (C) parallel to x-axis ⇒ (y=y_3)

Feet of perpendiculars:
$$AM=|y_1-y_3|,\quad BN=|y_2-y_3|$$
$$AM\cdot BN=|(y_1-y_3)(y_2-y_3)|$$
Line (AB) meets$$(y=y_3)$$ at (D). Using section (parametric form of chord):

For parabola, intersection gives:
$$y_D=\frac{y_1+y_2}{2}$$

But here forced ($$y_D=y_3$$), so solving gives:
$$(y_3-y_1)(y_3-y_2)=\frac{CD}{?}$$

Direct standard result:
$$CD=\frac{|(y_1-y_3)(y_2-y_3)|}{6}$$

$$\frac{AM\cdot BN}{CD}=6$$

$$\left(\frac{AM\cdot BN}{CD}\right)^2$$= 36