Construct the tangent to a point on the parabola
Matthew Barrera 
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.
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$\begingroup$Justification: Consider a general parabola $y=ax^2$. Don't need to consider the $bx+x$ term as these terms only describe horizontal and vertical translations but they don't change the parabola's shape. Consider point of tangency $P$ with coordinates $(t,at^2)$ and point $B$ follows $(0,at^2)$. From here your point $C$ is then $(0,-at^2)$. Now the slope of tangent $PC$ can be found with the basic slope formula "rise over run" using the points $P$ and $C$, which then becomes $m=\frac{at^2--at^2}{t-0}$ which simplifies to $2at$. This is the same value as when you take the derivative of $y=ax^2$ and plug in $x=t$. PS, this is a really neat construction, I am thinking of putting this question to my students...
$\endgroup$ 2 $\begingroup$Your supposition is correct.For parabola parametrized by $ (x,y)= (2 a t, a t^2)$ the slope at any point is $dy/dx=t$. It can be seen that tangent bisects the $x$ abscissa at tangent to vertex .. and the parabola vertex itself bisects (ordinate-central cutting point) line $ BOC.$
$\endgroup$ $\begingroup$Let point $C$ be such that $O$ is the midpoint of $BC$ on the parabola's axis.
I propose to take advantage of the interpretation of the parabola as a quadratic Bézier curve (). A basic property of the quadratic Bézier curve defined by $P,C,A$ ($P$ as origin, $A$ as endpoint, and $C$ as control point or "handle") is that the summit $O$ of the arc of parabola it defines is the midpoint of the midpoints of line segments $[CP]$ and $[CA]$.
The essential property of Bézier curve defined by $P,C,A$ in our case is that $CP$ is tangent to the curve in $P$.
The parametric equations of the curve can be given in the following form:
$$\binom{x}{y}=(1-t)^2\binom{x_P}{y_P}+2t(1-t)\binom{x_C}{y_C}+t^2\binom{x_A}{y_A}$$
for $0 \leq t \leq 1$. If $t=0$, we are in $P$. If $t=1$ we are in $A$.
It is important to check that the parabola passes through the midpoint $O$ of $BC$.
This will happen when $t=1/2$. In this case, the current point of the parabola is
$$\binom{x}{y}=\dfrac12\binom{x_C}{y_C}+(\dfrac14\binom{x_P}{y_P}+\dfrac14\binom{x_A}{y_A})=\dfrac12\binom{x_C}{y_C}+\dfrac12\binom{x_B}{y_B}$$
(using rules of barycentrical computations), i.e., the midpoint of line segment $CB$, that is to say point $O$, as desired.
$\endgroup$ 1 $\begingroup$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...
$\endgroup$ 2 $\begingroup$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.
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