How to find equation of normal to a conic? Find the equations of the three lines that pass through the point (9a,6a) and which are normal to the parabola y^2=4ax
 A: The parametric representation of the parabola is $x = at^2, y = 2at$. The normal at $t$ is given by $$y + tx = 2at + at^3$$ For the point $(9a,6a)$ to lie on this normal, we must have $$6a + 9at = 2at + at^3 $$ or $t^3 - 7t - 6 = 0$. This can be factored as $(t+2)(t+1)(t-3) = 0$. Hence either $t=-1, -2$ or 3. The normals are 
\begin{align*}
y-x &= -3a \\
y-2x &=-12a \\
y+3x &= 33a 
\end{align*} 
A: First, find an expression for the normal line to the graph at $(x_0,y_0)$.  Taking the derivative of $y^2=4ax$ implicitly gives $y' = \frac{2a}{y}$, so the equation of the normal line would be $$y - y_0 = -\frac{y_0}{2a}(x-x_0)$$
We know that the lines we are seeking all pass through the point $(9a,6a)$, so that point is a solution of the equation.
$$ 6a - y_0 = -\frac{y_0}{2a}(9a-x_0)$$
We also know that the point $(x_0,y_0)$ is a point on the graph of $y^2 = 4ax$, so $y_0^2 = 4ax_0$ and $x_0 = \frac{y_0^2}{4a}$.  Substituting gives
$$ 6a - y_0 = -\frac{y_0}{2a}\left ( 9a-\frac{y_0^2}{4a} \right)$$
Rewrite the equation into this form:
$$y_0^3 -28a^2y_0-48a^3 = 0$$
We know that the normal line through $(9a,6a)$ is one of the normal lines we are looking for, so $6a$ is a solution of the equation.  So use synthetic division to get $$(y_0-6a)(y_0^2+6ay_0+8a^2) = 0$$ and then factor $$(y_0-6a)(y_0+4a)(y_0+2a) = 0$$
So the three lines we are seeking are the normal lines at the points $(9a,6a)$, $(4a,-4a)$, and $(a,-2a)$.  Using the expression $-\frac{y}{2a}$ for the slope of the normal line, the slopes of the normal lines at those points would be -3, 2, and 1, respectively.  The lines are
$$y-6a=-3(x-9a)$$
$$y+4a=2(x-4a)$$
$$y+2a=(x-a)$$
