Geometry problem with parabola and two tangents which intersect 
Given a curve $P: x^2 = 8y$. Two tangents to $P$ are $L_1: y=m_1x+c_1$ and $L_2: y=m_2x+c_2$, and they are intersecting at a point $A$.
Problem 1. Express $c_1$ in terms of $m_1$


*

*Solution: $c_1 = - 2m_1 ^2$ $\leftarrow$  I don't know how $x$ is eliminated here.


Problem 2. Show that coordinates of $A$ are $(2(m_1+m_2), 2m_1m_2)$


*

*This reminds me of the angle between lines formula... but the angle is not given at this point, hence I don't know where to start.


Problem 3. If the angle between $L_1$ and $L_2$ is $\dfrac{\pi}{4}$, find the equation of the locus of $A$.


*

*Solution: $x^2-y^2-12y-4=0$ $\leftarrow$ Have no idea how to get here.


Anyone, please help, I am freaking out.
 A: First question:


*

*The line and the parabola are tangent therefore the system
$$
\left\{ \begin{gathered}
  x^2  = 8x \hfill \\
  y = m_1 x + c_1  \hfill \\ 
\end{gathered}  \right.
$$
must have two coincident solutions. The system get to you the following equation
$$
x^2  - 8m_1 x - 8c_1  = 0
$$
which has two coincident solution if and only if discriminant is zero. It means that
$$
64m_1 ^2  + 32c_1  = 0
$$
thus 
$$
c_1  =  - 2m_1 ^2 
$$
In the same way you get that
$$
c_2  =  - 2m_2 ^2 
$$
hence your lines are 
$$
\begin{gathered}
  L_1 :y = m_1 x - 2m_1 ^2  \hfill \\
  L_2 :y = m_2 x - 2m_2 ^2  \hfill \\ 
\end{gathered} 
$$
Second question 
The point A is the common point of the two lines thus it is the solution of the system
$$
\left\{ \begin{gathered}
  y = m_1 x - 2m_1 ^2  \hfill \\
  y = m_2 x - 2m_2 ^2  \hfill \\ 
\end{gathered}  \right.
$$
You have
$$
m_1 x - 2m_1 ^2  = m_2 x - 2m_2 ^2 
$$
which means
$$
x\left( {m_1  - m_2 } \right) = 2\left( {m_1 ^2  - m_2 ^2 } \right) = 2\left( {m_1  - m_2 } \right)\left( {m_1  + m_2 } \right)
$$
Since the lines are not parallel you have that $$m_1 \neq m_2$$ and therefore you have
$$
x = 2\left( {m_1  + m_2 } \right)
$$
By substitution you have
$$
y = 2m_1 \left( {m_1  + m_2 } \right) - 2m_1 ^2  = 2m_1 m_2 
$$
These are therefore the coordinates of A.
Third question
You have that
$$
\left| {\frac{{m_1  - m_2 }}
{{1 + m_1 m_2 }}} \right| = \tan \theta 
$$
In your case 
$$
\left| {\frac{{m_1  - m_2 }}
{{1 + m_1 m_2 }}} \right| = 1
$$
You can drop the absolute value by squaring:
$$
\left( {\frac{{m_1  - m_2 }}
{{1 + m_1 m_2 }}} \right)^2  = 1
$$
It follows that
$$
\left( {m_1  - m_2 } \right)^2  = \left( {1 + m_1 m_2 } \right)^2 
$$
and so
$$
m_1 ^2  + m_2 ^2  - 2m_1 m_2  = 1 + 2m_1 m_2  + \left( {m_1 m_2 } \right)^2 *
$$
Since
$$
m_1 ^2  + m_2 ^2  = \left( {m_1  + m_2 } \right)^2  - 2m_1 m_2 
$$
you can rewrite * as
$$
\left( {m_1  + m_2 } \right)^2  - 6m_1 m_2  = 1 + \left( {m_1 m_2 } \right)^2 
$$
Since, from question 2, you got that
$$
\left\{ \begin{gathered}
  m_1  + m_2  = \frac{{x_A }}
{2} \hfill \\
  m_1 m_2  = \frac{{y_A }}
{2} \hfill \\ 
\end{gathered}  \right.
$$
by substitution you get
$$
x_A^2  - y_A^2  - 12y_A  - 4 = 0
$$
A: I'm going to explain #1 which may help you figure the rest on your own.
$x^2=8y$
Differentiate both sides and isolate $y'$:
$$y' =\dfrac{x}{4}$$
Above is the slope of the tangent line to the parabola at $(x,y)$.

Next look at the first tangent line
$y = m_1x+c_1$
Say this tangent line touches the parabola at $(x_1,y_1)$, then the slope of the tangent line at this point is $m_1 = \dfrac{x_1}{4}$ 
Since the point $(x_1,y_1)$ is on both parabola and tangent line, it satisfies both:
$c_1 = y_1-m_1x_1 = \dfrac{{x_1}^2}{8} - m_1x_1 = \dfrac{(4m_1)^2}{8} - m_1(4m_1) = -2{m_1}^2$
A: The angular coefficent of the $L_1$ (knowing that the general coefficent $m=2ax_0+b$ for $ax^2+bx+c$) is: $m_1=\frac{1}{4}x_0$, so the line $L_1$: $y=\frac{1}{4}x_0x+c_1$. Now, the $\Delta$ of the following system has to be $0$: 
$$\left\{\begin{matrix}
y=\frac{1}{4}x_0x+c_1
\\ y=\frac{x^2}{8}
\end{matrix}\right.$$
Substituing and imposing $\Delta=0$, I have: $4x_0^2+32c_1=0$ so $c_1=-\frac{x_0^2}{8}$. Comparing $c$ and $m$, I have: $c_1=-m_1^2$.
