Problem statement:
Two tangents to the parabola $y=x^2$ are perpendicular. Prove that the intersection between the tangents lies on the line $y=-\frac{1}{4}$.
Solution:
Two tangents $y_1=k_1x+m$, $y_2=k_2x+n$ are perpendicular if $k_1k_2=-1$ so we can reduce the second tangent to $y_2=-\frac{1}{k_2}x+n$. Before finding the intersection which is $y_1=y_2$ I believe we need another condition that relates $y_1,y_2$ to $y=x^2$. I'm sure it has to do with the derivative but I am stuck.
Best Answer
Hints and facts here:
$y-y_0=2x_0(x-x_0)$ is taken as a line tangent to $f(x)$ at $(x_0,y_0)$.
$y-y_1=2x_1(x-x_1)$ is taken as a line tangent to $f(x)$ at $(x_1,y_1)$.
$y_1=x_1^2,~~y_0=x_0^2$.
$2x_1=(-1/2x_0)$.