Finding the equation of the normal to the parabola $y^2=4x$ that passes through $(9,6)$ 
Let $L$ be a normal to the parabola $y^2 = 4x$. If $L$ passes through the point $(9, 6)$, then $L$ is given by
(A) $\;y − x + 3 = 0$
(B) $\;y + 3x − 33 = 0$
(C) $\;y + x − 15 = 0$
(D) $\;y − 2x + 12 = 0$

My attempt: Let $(h,k)$ be the point on parabola where normal is to be found out. Taking derivative, I get the slope of the normal to be $\frac{-k}{2}$. Since the normal passes through $(9,6)$, so, the equation of the normal becomes:$$y-6=\frac{-k}{2}(x-9)$$$$\implies \frac{kx}{2}+y=\frac{9k}{2}+6$$
By putting $k$ as $2,-2,-4$ and $6$, I get normals mentioned in $A,B,C$ and $D$ above (not in that order).
But the answer is given as $A,B$ and $D$. What am I doing wrong?

The equation of the normal is $\color{red}{y=\frac{-kx}{2}+\frac{9k}{2}+6}$. The point $(h,k)$ lies on it so we get
$$k=\frac{-kh}{2}+\frac{9k}{2}+6.$$
But $k^2=4h$ (since $(h,k)$ lies on the parabola as well), so we get
$$k^3-28k-48=0 \implies (k-6)(k+2)(k+4)=0 \implies  k=6,-2,-4.$$
These value can be used to find the equation of the normal as:
\begin{align*}
y&=-3x+33\\
y &=x-3\\
y &=2x-12
\end{align*}
