Construct the tangent to a point on the parabola Problem: 
Given the following parabola, construct a tangent to point $P$. Justify the construction.

Solution: 
Draw the line $PA$ which is perpendicular to the axis and intersects that axis at $B$. Mark off the distance $OB$ and use it to find point $C$. The line $PC$ is tangent to the parabola.

Basically my only justification that CP is tangent is that it looks really tangent. I have tried justifying my claim but can't seem to get anywhere. I have noticed that a circle can be constructed through points $PAC$. Let point $D$ be where $\circ PAC$ intersects the axis of the parabola. Then $\triangle CPD$ is an inscribed right triangle. This fact seems like it could be useful, but I'm not sure.

Take a parabola: $$y=ax^2+bx+c$$
Differentiating gives us the slope equation: $$m=2ax+b$$
The slope is zero at the vertex so the $x$ value of the vertex is: $$x=\frac{-b}{2a}$$
Sub this into the parabola equation to get the $y$ value at the vertex. The vertex is: $$(\frac{-b}{2a},c-\frac{b^2}{4a})$$
Let's pick a point on the parabola $k$ above the vertex. So its $y$ value is: $$y=c-\frac{b^2}{4a}+k$$
Let's prove that the tangent intersects the axis of symmetry at: $$c-\frac{b^2}{4a}-k$$
Let's find the two $x$ values at: $y=c-\frac{b^2}{4a}+k$
Sub this value for $y$ into the parabola equation and use the quadratic formula to get your two values for $x$. $$x=\frac{-b}{2a}\pm \sqrt{\frac{k}{a}}$$
Let's use the one with the positive sign.
The point of tangency is : $$(\frac{-b}{2a}+\sqrt{\frac{k}{a}},c-\frac{b^2}{4a}+k)$$
Sub this $x$ value into the slope equation to get the slope of the tangent. It is: $$m=2\sqrt{ak}$$
Using the value for the slope and the point of tangency let's derive the equation of the tangent. $$y-c+\frac{b^2}{4a}-k=2\sqrt{ak}(x+\frac{b}{2a}-\sqrt{\frac{k}{a}})$$
We want to see where this tangent line intersects the axis. The axis is the line $x=\frac{-b}{2a}$. Sub $x=\frac{-b}{2a}$ into the tangent equation and solve for $y$.
$$y=c-\frac{b^2}{4a}-k$$
Q.E.D.
