Prove that only one normal to the parabola $y^2=4(x-11)$ passes through the focus $(12,0)$

Question

question on the title, thanks!! I think it has to do with the normal gradient equation, which i believe is $y-y^*=-\frac y2(x-x^*)$ I have no clue what to do next. :(

Answer 1

Since $y_1-y=-\frac{y_1}{2}(x_1-x)$, letting $y=0$ and $x=12$ gives

$2y_1=-y_1(x_1-12)$ where $y_1^2=4(x_1-11)$, so $x_1=\frac{1}{4}y_1^2+11$.

Now substitute for $x_1$, and find the solutions for $y_1$.