We start with a breakdown of the problem. We will assume that the tangent lines intersect at some point #x#, and that they are tangent to the parabola at the points #(x_1, y_1)# and #(x_2,y_2)#. What this problem is really asking is for us to prove that the point of intersection is the midpoint of the points of tangency. We can see this illustrated visually in the picture.

The point of intersection is x=1, and the points of tangency are x=0 and x=2. And what do you know! The midpoint of 0 and 2 is 1; in other words, *the lines intersect at the midpoint of the points of tangency*.

Calculus tells us that the slope of the parabola #ax^2# is #2ax#. If you are not familiar with the concept of derivative, it basically means that at any point #x#, the parabola's slope (rate of change) is #2ax#. The reason why we need slope is to express the equation of the tangent lines:

#y-y_1=m_1(x-x_1)#

#y-y_2=m_2(x-x_2)#

are our two tangent lines. (This comes from the general formula for equation of a straight line).

Now, we know #y_1=a(x_1)^2# and #y_2=a(x_2)^2#, simply by plugging in #x_1#. We also know that #m# (slope) #=2ax#, so #m_1=2ax_1# and #m_2=2ax_2#. Plugging into the above equations, we have:

#y-a(x_1)^2=2ax_1(x-x_1)#

#y-a(x_2)^2=2ax_2(x-x_2)#

To make life easier, we will make #c=x_1# and #d=x_2#, so it's easier on the eyes:

#y-ac^2=2ac(x-c)#

#y-ad^2=2ad(x-d)#

Simplifying the equations yield:

#y=2acx-2ac^2+ac^2#

#y=2adx-2ad^2+ad^2#

And finally:

#y=2acx-ac^2#

#y=2adx-ad^2#

I want to briefly explain the significance of this result. Now we have the general equation for two lines tangent to a parabola at #x=c# and #x=d#. To find where these two lines intersect, we set these equations equal to each other:

#2acx-ac^2=2adx-ad^2#

And solve for the point of intersection #x#:

#2acx-ac^2=2adx-ad^2#

#2acx-2adx=ac^2-ad^2#

#x(2ac-2ad)=ac^2-ad^2#

#x=(ac^2-ad^2)/(2ac-2ad)#

Here's the fun part (though it does require a bit of factoring):

#x=(ac^2-ad^2)/(2ac-2ad)#

#x=(a(c^2-d^2))/(2a(c-d))#

#x=((c-d)(c+d))/(2(c-d))#

#x=(c+d)/(2)#

Remember how #c=x_1# and #d=x_2#? Making those substitutions yields:

#x=(x_1+x_2)/(2)#

Which means the point of intersection (#x#) is the midpoint of the points of tangency, which is what we wanted to prove.