Intersection of two tangents on a parabola proof There are two tangent lines on a parabola $x^2$. The $x$ values of where the tangent lines intersect with the parabola are $a$ and $b$ respectively. The point where the two tangent lines intersect has an $x$ value of $c$. 
Prove that $c=(a+b)/2$
I have tried taking the derivative of $x^2$ and using the point slope form to find each tangent line's equation, and then making each tangent equal to each other to find that $c$ is equal to $a+b$, but each time I try it out I can't get anything out of the algebra

Let's try to work through that algebra, then.
The derivative of $x^2$ is $2x$, so those tangent lines are given by
$$
y = 2a(x-a) + a^2\\
y = 2b(x-b) + b^2
$$
Thus, at the $x$-coordinate of the lines' intersection, we have
$$
2a(x-a) + a^2 = 2b(x-b) + b^2
$$
Expanding the above expressions, we have
$$
2ax - 2a^2 + a^2 = 2bx - 2b^2 + b^2\implies\\
2ax - a^2 = 2bx - b^2
$$
The tricky part comes next, where we factor in order to "combine like terms". We have
$$
2ax - 2bx = a^2 - b^2\\
2(a - b)x = a^2 - b^2
$$
Now, assuming $a\neq b$, we may divide both sides by $2(a-b)$ to get
$$
x = \frac 12 \frac{a^2 - b^2}{a - b}
= \frac 12 \frac{(a-b)(a+b)}{a - b}
= \frac 12 (a+b)
$$
As desired. I hope that clears things up.
