Intersection of two tangents on a parabola proof

Question

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

Answer 1

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.

Answer 2

Consider the curve $C$ with the equation $y=x^2$. When $x=a$, we have $y=a^2$. The derivative is given by $y'=2x$ and so the gradient at the is exactly $2a$. The equation of the tangent line is

$$y-a^2=2a(x-a)$$

Expanding this gives $y-2ax=-a^2$. Similarily, the equation of the tangent line to $C$ at the point with $x$-coordinate $x=b$ is given by $y-2bx=-b^2$. We are left needing to solve

\begin{eqnarray*} 2ax-y&=&a^2 \\ 2bx-y&=&b^2 \end{eqnarray*} Subtracting the second equation from the first gives $2(a-b)x=a^2-b^2$. Notice that $a^2-b^2$ factorises to give $(a-b)(a+b)$ and hence: $2(a-b)x=(a-b)(a+b)$. Assuming $a \neq b$ we can divide both sides by $a-b$ to give $2x=a+b$ or, better still $x=\tfrac{1}{2}(a+b).$