Problem source

Find the equations of the tangent and normal to the parabola $y^{2}=4ax$ at the point $(at^{2},2at).$
For $i=1,2,$ and 3, let $P_{i}$ be the point $(at_{i}^{2},2at_{i}),$ where $t_{1},t_{2}$ and $t_{3}$ are all distinct. Let $A_{1}$ be the area of the triangle formed by the tangents at $P_{1},P_{2}$ and $P_{3},$ and let $A_{2}$ be the area of the triangle formed by the normals at $P_{1},P_{2}$ and $P_{3}.$ Using the fact that the area of the triangle with vertices at $(x_{1},y_{1}),(x_{2},y_{2})$ and $(x_{3},y_{3})$ is the absolute value of 
\[
\tfrac{1}{2}\det\begin{pmatrix}x_{1} & y_{1} & 1\\
x_{2} & y_{2} & 1\\
x_{3} & y_{3} & 1
\end{pmatrix},
\]
show that $A_{3}=(t_{1}+t_{2}+t_{3})^{2}A_{1}.$
Deduce a necessary and sufficient condition in terms of $t_{1},t_{2}$ and $t_{3}$ for the normals at $P_{1},P_{2}$ and $P_{3}$ to be concurrent.

Solution source

$\frac{dy}{dt} = 2a, \frac{dx}{dt} = 2at \Rightarrow \frac{dy}{dx} = \frac{1}{t}$.

Therefore the equation of the tangent will be $\frac{y - 2at}{x-at^2} = \frac{1}{t} \Rightarrow y = \frac1tx +at$ and normal will be $\frac{y-2at}{x-at^2} = -t \Rightarrow y = t(at^2-x+2a)$.

The tangents will meet when:
\begin{align*}
&& \begin{cases} t_iy -x &= at_i^2 \\
t_j y - x &= at_j^2 \\
\end{cases} \\
\Rightarrow &&(t_i - t_j)y &= a(t_i-t_j)(t_i+t_j) \\
\Rightarrow && y &= a(t_i+t_j) \\
&& x &= at_it_j
\end{align*}

The normals will meet when:
\begin{align*}
&& \begin{cases} y +t_i x &= at_i^3+2at_i \\
y +t_j x &= at_j^3+2at_j \\
\end{cases} \\
\Rightarrow &&(t_i - t_j)x &= a(t_i-t_j)(t_i^2+t_it_j+t_j^2+2) \\
\Rightarrow && x&= a(t_i^2+t_it_j+t_j^2+2) \\
&& y &= -at_it_j(t_i+t_j)
\end{align*}

Therefore the area of our triangles will be:

\begin{align*}
\tfrac{1}{2}\det\begin{pmatrix}at_1t_2 & a(t_1+t_2) & 1\\
at_2t_3 & a(t_2+t_3) & 1\\
at_3t_1 & a(t_3+t_1) & 1
\end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\
t_2t_3 & (t_2+t_3) & 1\\
t_3t_1 & (t_3+t_1) & 1
\end{pmatrix} \\
&= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\
t_2(t_3-t_1) & (t_3-t_1) & 0\\
t_1(t_3-t_2) & (t_3-t_2) & 0
\end{pmatrix} \\
&= \frac{a^2}{2}|(t_2-t_1)(t_3-t_2)(t_1-t_3)|
\end{align*}
and
\begin{align*}
\tfrac{1}{2}\det\begin{pmatrix}a(t_1^2+t_1t_2+t_2^2+2) & -at_1t_2(t_1+t_2) & 1\\
a(t_2^2+t_2t_3+t_3^2+2) & -at_2t_3(t_2+t_3) & 1\\
a(t_3^2+t_3t_1+t_1^2+2) & -at_3t_1(t_3+t_1) & 1\\
\end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\
(t_2^2+t_2t_3+t_3^2+2) & -t_2t_3(t_2+t_3) & 1\\
(t_3^2+t_3t_1+t_1^2+2) & -t_3t_1(t_3+t_1) & 1\\
\end{pmatrix} \\
&= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\
t_3^2-t_1^2+t_2(t_3-t_1) & t_2(t_1^2+t_1t_2-t_2t_3-t_3^2) & 0\\
t_3^2-t_2^2+t_1(t_3-t_2) & t_1(t_2^2+t_2t_1-t_1t_3-t_3^2) & 0\\
\end{pmatrix} \\
&= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\
(t_3-t_1)(t_3+t_2+t_1) & t_2(t_1-t_3)(t_1+t_3+t_2) & 0\\
(t_3-t_2)(t_3+t_2+t_1) & t_1(t_2-t_3)(t_1+t_2+t_3)& 0\\
\end{pmatrix} \\
&= \frac{a^2}{2}(t_1+t_2+t_3)^2|(t_2-t_1)(t_3-t_2)(t_1-t_3)|
\end{align*}

as required.

The normals will be concurrent iff the area of their triangle is $0$. This is certainly true if $t_1+t_2+t_3 = 0$. In fact the only if is also true, since no $3$ tangents can be concurrent.