Maximum of three normals pass through point inside parabola proof (1 Viewer)

FootballRobotic · Nov 29, 2016

RTP: From any point located inside the parabola, a maximum of three normals pass through it.

braintic · Nov 30, 2016

The equation of the normal is cubic in t. Hence it has up to three real roots (dependent on the value of the point [x,y] you substitute into the normal equation), each root (value of t) corresponding to a point on the parabola.

Lugia101101 · Dec 1, 2016

A slightly more "rigorous" proof (not really):
$\text{Take the general form of the parabola }x^2=4ay\text{, and the points }P(2ap,ap^2)\text{ and }A(x_0,y_0)\text{.}$
$\text{For a parabola with a vertex at }(h,k)\text{, just shift }A\text{ to an equivalent point.}$
$\begin{align*}\text{So }x^2&=4ay\\ y&=\frac{x^2}{4a}\\ \frac{dy}{dx}&=\frac{x}{2a}\\ -\frac{dx}{dy}&=-\frac{2a}{x}\end{align*}$
$\,\!\!\!\!\!\!\!\!\!\!\text{At }P\text{, }x=2ap\text{, so }-\frac{dx}{dy}=-\frac{2a}{2ap}=-\frac{1}{p}\\ \text{In order to find all points that satisfy the condition, we must find }p\text{ such that:}\\ \begin{align*}-\frac{dx}{dy}&=AP_m\\ -\frac{1}{p}&=\frac{y_0-ap^2}{x_0-2ap}\\ 2ap-x_0&=y_0p-ap^3\\ 0&=ap^3+(2a-y_0)p-x_0\end{align*}$
$\text{So let }q=ap^3+(2a-y_0)p-x_0$
$\text{Now this function }q\text{ in }p\text{ describes the difference in gradient between the normal at }P\text{ and the line }AP\text{.}$
$\text{For them to be the same, }q=0\text{, but as this is a cubic, there will be 1, 2, or 3 solutions for }p\text{.}$
$\text{For the proof to be valid, we must prove there is at least one point }A\text{ that lies within the parabola and produces a cubic }q\text{ with three distinct solutions.}$
$\text{If two stationary points exist, both of which are non-zero, and lie on opposite sides of the q-axis, then there are three distinct roots to the polynomial }q\text{.}$
$\text{So }\frac{dq}{dp}=3ap^2+2a-y_0$
$\begin{align*}\text{And }0&=3ap^2+2a-y_0\text{ for the stationary points' p-coordinates.}\\ 3ap^2&=y_0-2a\\ p^2&=\frac{y_0-2a}{3a}\\ p&=\pm\sqrt{\frac{y_0-2a}{3a}}\end{align*}$
$\text{So, we can now test a point. Say, }A(0, 5a)\text{:}$
$\text{The stationary points are at: }p=\pm\sqrt{\frac{5a-2a}{3a}}=\pm1$
$\begin{align*}\text{So, }q&=a(1)^3+(2a+5a)(1)-0,a(-1)^3+(2a+5a)(-1)-0\\ &=8a, -8a\end{align*}$
$\text{So as }a\neq0\text{, the stationary points are non-zero and lie on opposite sides of the q-axis.}$
$\therefore\text{There exists at least one }A\text{ that lies within the parabola such that three normals can be drawn through it.}$
$\text{As }q\text{ is a polynomial in }p\text{ of degree three, there exist a maximum of three real roots.}$
$\text{Hence, for any point within a parabola, a maximum of three normals can be drawn through it.}$

Maximum of three normals pass through point inside parabola proof (1 Viewer)

FootballRobotic

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braintic

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Lugia101101

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