Tangents to a parabola that go through the same point The question is: The two lines tangent to f (x) = $x^2$ + 4x + 2 through the point (2, -12)have equations y = ax + b and y = cx + d, respectively. What is the value of a + b + c + d?
What I did to solve it:
f '(x) = 2x + 4. The point is (2 -12) so I plugged in two to get f '(2) = 8. I used this in point-slope form to get y + 12 = 8(x-2) => y = 8x -28. Since it is two equations, I added them up and multiplied by 2 and got -40, which is the answer. I feel like this is a fluke and it doesn't make sense to me why this worked. Can someone explain either why it works or a way that will always work? Thanks in advance!

Look at the figure:

The point $A=(2,-12)$ is not a point of the parabola, so the slopes of the tangents to the parabola from this point cannot be be simply derived starting from  the derivative of the parabola at $x=2$.
If $P=(X,X^2+4X+2)$ is a point of tangency (the points $C$ and $D$ in the figure), than the slope $m=(y_P-y_A)/(x_P-x_A)$ of the line $PA$ must be the same as the slope of the tangent to the parabola in $P$, i.e.: $m=f'(x_P)$ . So, For our $A$ and $P$ we have:
$$
\frac{X^2+4X+2+12}{X-2}=2X+4
$$ 
Solving this equation we find the coordinates $x_C$, $x_D$ of the two points of tangency, from wich we can find the equations of the two tangent lines.
Another simple solution, without using derivative, is to note that the system:
$$
\begin {cases}
y=x^2+4x+2\\
y+12=m(x-2)
\end{cases}
$$
represents the intersection between the parabola and the lines that passe thorough $A$, and a line is tangent  to the parabola if  the system has only one (double) solution.  This means that the discriminat $\Delta(m)$  of the second degree equation  that solve the system is such that: $\Delta(m)=0$ . 
Since $\Delta(m)$ is a second degree polynomial in the parameter $m$, this is a second degree equation in $m$ that gives the two values of $m$ for the slopes for the two tangent lines.
