Normal Equations to Parabola Passing Through Given Point

To find the equations of the normal lines $L$ to the parabola $y^2 = 4x$ passing through the point $(9, 6)$, we use the slope-intercept form of the normal to a parabola.

The standard equation of the parabola is $y^2 = 4ax$. Here, $4a = 4 \implies a = 1$.

The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by:
$y = mx - 2am - am^3$

Substituting $a = 1$:
$y = mx - 2m - m^3$

Since the normal $L$ passes through the point $(9, 6)$, we substitute $x = 9$ and $y = 6$ into the equation:
$6 = m(9) - 2m - m^3$
$6 = 7m - m^3$
$m^3 - 7m + 6 = 0$

Now, we solve this cubic polynomial for $m$:
Observing that $m = 1$ is a root ($1^3 - 7(1) + 6 = 0$), we factor the cubic expression:
$(m - 1)(m^2 + m - 6) = 0$
$(m - 1)(m + 3)(m - 2) = 0$

Thus, the possible values for the slope $m$ are:
$m = 1, \quad m = -3, \quad m = 2$

Now, we determine the equation of the normal line for each value of $m$:

1. For $m = 1$:
   $y = (1)x - 2(1) - (1)^3$
   $y = x - 3 \implies y - x + 3 = 0$
   This corresponds to Option (A).

2. For $m = -3$:
   $y = (-3)x - 2(-3) - (-3)^3$
   $y = -3x + 6 + 27$
   $y = -3x + 33 \implies y + 3x - 33 = 0$
   This corresponds to Option (B).

3. For $m = 2$:
   $y = (2)x - 2(2) - (2)^3$
   $y = 2x - 4 - 8$
   $y = 2x - 12 \implies y - 2x + 12 = 0$
   This corresponds to Option (D).

Hence, the correct options are A, B, and D.