The normal at $T(at^2,2at)$ of parabola $y^2=4ax$ meets the parabola again at $S(as^2,2as)$. Show that $t^2+st+2=0$. 
The normal at the point $T(at^2,2at), t\not = 0$, on the parabola $y^2=4ax$ meets the parabola again at the point $S(as^2,2as)$. Show that $t^2+st+2=0$.

I am completely lost. I tried using implicit differentiation but that isn't working, so I really don't know what to do...
 A: The equation of the parabola is $C:y^2=4ax$
So, $2y\frac{dy}{dx} = 4a$ or $\frac{dx}{dy} = \frac{y}{2a}$
The slope of the normal at $T$ is $m_N =-\frac{dx}{dy}\big|_T = -\frac{2at}{2a} = -t$
Also, the slope has a formula of $m_N = \frac{y_2-y_1}{x_2-x_1}$
As, T and S both lie on the normal, 
$m_N = \frac{2as-2at}{as^2-at^2} = 2\frac{s-t}{s^2-t^2} = 2\frac{1}{s+t}$ (if $s\not=t$)
So, $-t = 2\frac{1}{s+t}$
$-st -t^2 = 2$
or$$t^2+st+2= 0$$
A: Thinking in vectors, you could work this way:


*

*The tangent vector at $T$ is $\frac{dT}{dt}=(2 a t, 2 a )$, which we can divide-through by $2a$ to get $(t,1)$.

*The normal vector $\overrightarrow{ST}$ is $T-S = (2a(t^2-s^2),2a(t-s))$, which (for $s\neq t$) we can divide-through by $2a(t-s)$ to get $(t+s,1)$.

*Since the tangent and normal vectors are perpendicular, their dot product vanishes:
$$ 0 = (t,1)\cdot(t+s,1) = t(t+s)+1 = t^2 + ts + 1 \qquad\square$$

Of course, this is completely equivalent to working with slopes, but this way avoids fractions.
A: With $m_T = \frac{2a}{y_T}$ the normal line at $T$ passing by $S$ is
$$
y_S-y_T = -\frac{1}{m_T}(x_S-x_T)
$$
or
$$
y_S-y_T = -\frac{1}{m_T}\left(\frac{y_S^2}{4a}-\frac{y_S^2}{4a}\right)
$$
or
$$
1 = -\frac{y_T}{8a^2}(y_S+y_T)\Rightarrow 8a^2 = -2at(2as+2at)
$$
A: A different (parabolic) approach.  Given any secant $TS$ of the parabola, there is exactly one point $M$ of the parabola in which the tangent is parallel to the secant. Furthermore, 

the second coordinate of $M$ is the mean of the second coordinates of $T$ and $S$,

a (hopefully) known property of parabolas.
Since $y'=2a/y$, the normal's slope in $T$ is $-t$, as already pointed out.  Setting $2a/y_M=-t$ we get $y_M=-2a/t$.  We know that 
$$y_M=\frac{y_T+y_S}{2},$$
from where $y_S=2a(-2/t-t)$, hence $s=-2/t-t$.  This gives $t^2+st+2=0$.
