Construct the tangent to a point on the parabola

Question

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.

Answer 1

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...

Answer 2

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.$

Answer 3

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...

Answer 4

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 (https://www.futurelearn.com/courses/maths-linear-quadratic-relations/0/steps/12129). 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.

Answer 5

I think the most reasonable geometric justification would be to show that any ray coming parallel to the axis will reflect towards the focus, i.e. that the normal is the bisectriz of the angle EAF

For this it should be enough to check consistency of the construction. Perhaps the most important point is that FC is the same length that AF. This follows from the definition of parabola, because AF = AA' = DF' and then just by reflection in the origin DF' = FC.

From this, taking the circle centered in the focus, we build B and thus the normal AB. This is actually my common way to do the tangent and normal, but if only need the tangent, projection is fine, less clutter.

Ah, we have still not proven that angle BAF is half of EAF. Well, honestly I do not know what argument is more intuitive. We could just parallel translate AF pointing its arrow to B, and see that as AF=BF we have now a rombus, call it (AFBE'). That would be enough to my taste.

https://www.geogebra.org/m/h4295ewq