# Intersection of two tangents on a parabola proof

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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.

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### Norwolf

Just a university student.

Updated on November 21, 2020

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