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.

EternusVia. Here is another, but NON calculus approach: Consider point of tangency $P(t,at^2)$. Tangent PC passing through P is then of the form $y=mx+at^2-mt $ (verify!).Now intersecting the tangent with the parabola gives us the equation $ax^2=mx+at^2-mt,$ which simplifies to $ax^2-mx-at^2+mt=0.$ This equation is to have only one solution (at $P$) and so the Discriminant $D=b^2-4ac,$ of this equation must be zero. This gives $m^2-4a(-at^2+mt)=0$. Simplify gives $m^2-4amt+4a^2t^2=0$. Surprise...surprise, this equation is factorable as a perfect square: $(m-2at)^2=0$ from which $m=2at$ , etc...
