Let $F$ be the focus of the parabola, $HG$ its directrix, with vertex $V$ the midpoint of $FH$. From the definition of parabola it follows that $PF=PG$, where $P$ is any point on the parabola and $G$ its projection on the directrix.
The tangent at $P$ is the angle bisector of $\angle FPG$, hence it is perpendicular to the base $GF$ of isosceles triangle $PFG$, and intersects it at its midpoint $M$.
But the tangent at $V$ is parallel to the directrix and bisects $FH$, hence it also bisects $FG$ at $M$, as it was to be proved.
Some comments were really helpful but I believe none gave me the intuition for this, which is what I really didn't understand , however I'm truly grateful for everyone that commented or attempted to help me.
Here's my approach after dreaming about it the entire night. (Not just the intuition)
We will first determine a general equation for all lines tangent to the function $f(x)$ at any given x , to do so let's create a generic point for our lines, let's make the point $(a,f(a))$ or equivalently $(a,a^2)$
We proceed to get the slope at any given value by getting the $d/dx$
of our function $f(x)$ wich results to be $f'(x) = 2x$
Next, let's evaluate the slope at our generalized point $(a,a^2)$
$$f'(a) = 2a$$
By doing this we can start to construct our line equation with $2a$ as the slope.
$$y=2ax+b$$
We're missing the b value, and to solve for it we will use the point that we are sure lie on our line equation, that is our point $(a,a^2)$ , so let's substitute it into our equation.
$$a^2=2a(a)+b$$
which is equal to
$$a^2=2a^2+b$$
Next, let's move the non b terms to the left , resulting
$$-a^2=b$$
With that we can substitute our b to our first equation, resulting
$$y=2ax-a^2$$
This is the tricky part, we need to find which values of a would contain the point (1,0) (On this specific example) so we can substitute x = 1 and y = 0 on our equation, resulting
$$0=2a(1)-a^2$$
equivalently
$$0=2a-a^2$$
We proceed to factor out the a as
$$0=a(2-a)$$
We know that if our outcome it's zero then one of the terms being multiplied is going to be equal to zero, so we can say that either
$$a=0$$
or the term
$$2-a=0$$
Which we can easily solve for a resulting in
$$-a=-2$$
or $$a=2$$
so either a is equal to zero or a is equal to 2 , so we can substitute our "a's" on our first defined function for all tangent lines $y=2ax-a^2$, resulting in the line equations
$$y=2(2)x-(2^2)$$
Which is equivalent to $$y=4x-4$$
and $$y=2(0)x-(0^2)$$
Which is equivalent to $$y=0$$
So by doing these things we managed to get the equation of the lines tangent to $f(x)=x^2$ that go through the point (1,0)
I hope that this helps someone!
Regards.
Best Answer
From the tangent line, we see that the parabola has y'(0)=3 at (0,6). This gives: $$y'(0)=2A(0)+B=3$$$$B=3$$ From y(0)=6, we get: $$y(0)=A(0)^2+B(0)+C=6$$ $$C=6$$ From y(1)=0, we get: $$y(1)=A(1)^2+3(1)+6=0$$ $$A+9=0$$ $$A=-9$$