Intersection of two tangents on a parabola proof
Let's try to work through that algebra, then.
The derivative of $x^2$ is $2x$, so those tangent lines are given by $$ y = 2a(xa) + a^2\\ y = 2b(xb) + b^2 $$ Thus, at the $x$coordinate of the lines' intersection, we have $$ 2a(xa) + a^2 = 2b(xb) + b^2 $$ 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 $2(ab)$ to get $$ x = \frac 12 \frac{a^2  b^2}{a  b} = \frac 12 \frac{(ab)(a+b)}{a  b} = \frac 12 (a+b) $$ As desired. I hope that clears things up.
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Norwolf almost 3 years
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