Intersection of two tangents to a parabola 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.
 A: 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 $(x_1,y_1)$.

*$y_1=x_1^2,~~y_0=x_0^2$.

*$2x_1=(-1/2x_0)$.
A: Step 1:   The tangent to the parabola at the point $x_0$ has the form
$$f(x)=2xx_0-x_0^2$$
(you know the slope by differentiation and one point on the tangent, namely $(x_0,x_0^2)$).
Step 2: The tangents at $x_0$ and $x_1$ will intersect at $(\frac{x_0+x_1}{2},x_0x_1)$
A: The tangents to the points $(x_1,x_1^2)$, $(x_2,x_2^2)$, with $x_1,x_2\neq 0$, $x_1\neq x_2$ are
$$y =2x_i x -x^2_i, $$
for $i=1,2$. In particular, the slope is found using $y'(x)=2x$.
As the 2 tangents are supposed to be perpendicular, then
$$4x_1x_2=-1, $$
or $x_1x_2=-\frac{1}{4}$. All you need is to find the point of intersection between the lines: subtract $y =2x_2 x -x^2_2 $ from $y =2x_1 x -x^2_1 $ arriving at
$$x=\frac{1}{2}(x_2+x_1). $$
Plug such coordinate into $y =2x_2 x -x^2_2 $ arriving at
$$y=x_2(x_2+x_1)-x^2_2$$
or
$$y=x_2x_1=-\frac{1}{4}$$
as claimed.
A: A parabola can be defined as the set of all points that have the same distance from a point (the focus) and a line (the directrix).
Note that $y = -\frac 1 4$ is the directrix of the parabola $y = x^2$. So we are to prove that the two tangents intersect on the directrix.
A short Lemma:
Let $g$ be the directrix line and $F$ the focus of the parabola. 
Let $l$ be a tangent to the parabola and let $P$ be the point where it intersects the parabola.  Let $A$ the orthogonal projection of $P$ onto $g$. Then it is easy to see from above definition of a parabola that $l$ is the perpendicular bisector of the line segment $AF$ (simply because this perpendicular bisector intersects the parabola at the point $P$ and in no other point). (End of Lemma)
In our case we have two tangents $l_1$ and $l_2$ which are perpendicular. Let $A_1$ and $A_2$ be the corresponding points on $g$ (as in the lemma). Then $A_1 F$ and $A_2 F$ must also be perpendicular. So the angle $A_1 F A_2$ is a right angle. Therefore the circumcenter of the triangle $A_1 F A_2$ (which is where the lines $l_1$ and $l_2$ intersect) is on the line segment $A_1 A_2$ (which is part of the line $g$.)
