# Interesting rotating parabola proof

So, I am reading linear algebra and cartesian geometry by Juan de Burgos for the first time, and came across this problem that I just can't figure out. Goes like this:

Consider a parabola, whose distance from the focus to the directrix is a fixed p, and that rotates while keeping its focus fixed. We also consider a certain fixed direction in the parabola's plane, and consider the tangent to the parabola that has that given direction. Find the equation of the geometric space that all the tangency points describe.

The original question is in Spanish but I have tried to translate it the best I can. If anyone speaks Spanish I can send you the original question.

Edit: Ill keep working on it and try to write my work in here, sorry that I didnt do from the beginning

Ok, so first off the book gives the soultion to a problem in the back, but without any explanation or none of that. it says:

**Origin at the focal point, << x axis>> in the given direction;

directrix $$xcos(\theta)+ysin(\theta)=p$$.

Parabola $$x^2 sin^2(\theta)+y^2 cos^2(\theta)-2xysin(\theta)cos(\theta)+2p(xcos(\theta)+ysin(\theta))=p^2$$

The polar of (0,1,0) is $$xsen^2(\theta)-ysin(\theta)cos(\theta)+pcos(\theta)=0$$.

Eliminating $$\theta$$ we get $$p^2(x^2+y^2)=4y^4$$.**

I get the part of creating a new coordinates taking the focal point to be F(0,0). as so: see image here The directrix in a standard parabola where the vertix is in the origin has directrix x=p/2, so because we are taking the focal point it would be now x=p. Because of the rotation described for the parabola, i thought of using rotation of coordinate axes, as so: see image And thats why the directrix would be $$xcos(\theta)+ysin(\theta)=p$$ as described in the solution given by the book. (Thanks to a commenter here who clarified to me that $$xcos(\theta)+ysin(\theta)=p$$ or $$xcos(\theta)+ysin(\theta)=-p$$ would make no difference, I was confused on that too).

Now, how in the world did they get the parabola described in the solution and the stuff that comes after?

I have read the chapter like three times but I dont know why i cant understand this one problem

• So is the question simply, how do we know the equation of a line at angle $\theta$ and distance $p$ from the origin is $x\cos(\theta)+y\sin(\theta)=p$? – David K Dec 4 '20 at 14:33
• You can check that the given directrix has distance $p$ from the origin. – Intelligenti pauca Dec 4 '20 at 14:34
• @DavidK no, the question is to determine what geometric space the tangency points describe as the parabola moves. what I asked is only the first step – sara 15 Dec 4 '20 at 16:03
• "how in the world did they get the parabola described in the solution" ... It's hard to say for sure what the author's intent is without knowing the topic(s) covered in the chapter. However, since the solution starts with the directrix equation, you can use the focus-directrix definition of the parabola: "dist. from focus = dist. from directrix". For the given focus and directrix, this means $$\sqrt{(x-0)^2+(y-0)^2} = \frac{|x\cos\theta+y\sin\theta-p|}{\sqrt{\cos^2\theta+\sin^2\theta}}$$ Squaring and simplifying gives the parabola equation (parts of which are missing in your question :). – Blue Dec 5 '20 at 6:03
• @blue Im pretty certain this is exactly what I need, will try it right now. Also, I didnt know the word locus as they never call it that in my book, but I googled it and am familiar with the term now, so thankyou very much – sara 15 Dec 5 '20 at 7:05

## 2 Answers

HINT.

Instead of following the path suggested by the book, you could have the parabola fixed and rotate the direction of the tangent: it is not difficult to find a function $$r=f(\phi)$$, where $$r=FP$$ is the distance between focus $$F$$ and tangency point $$P$$, while $$\phi$$ is the angle between line $$FP$$ and the tangent at $$P$$.

It may also be useful to remember that the tangent line at $$P$$ is the bisector of $$\angle FPH$$, where $$H$$ is the projection of $$P$$ on the directrix.

I understood the problem in this way:

A parabola (focus to directrix distance = 2p) rotates about its focus at origin. Find equation to the variably rotated parabola and its tangent as function of the rotation angle $$\alpha$$.

Answer:

I have tried to rotate the entire parabola/tangent at vertex as one combination.

The given parabola equation:

$$x^2 \sin ^2 t + y^2 \cos^2 t-2 x y \sin t \cos t+y^2 \cos ^2 t +2 p (x \cos t +y \sin t) = p^2 \tag 1$$

For $$t=0$$ it reduces to $$2 p x + y^2 = p^2 \tag 2$$

Bring in parameter $$t$$ in a slightly different way

$$x= \dfrac{p^2-t^2}{2p}, y = t \tag 3$$

and then we try to rotate the entire parabola through $$\alpha$$.

$$x\to x \cos \alpha- y \sin \alpha , y\to x \sin \alpha +y \cos \alpha \tag 4$$

Solving for $$(x,y)$$ from the rotated coordinates ( rather than directly) with new parametrization $$t$$ included

$$x \cos \alpha- y \sin \alpha =t, \; x \sin \alpha +y \cos \alpha = \dfrac{p^2-t^2}{2p} \tag 5$$

Next put $$t=0$$ gives starting point $$( x=0, y=p/2)$$ that is the point of tangency of tangent $$x= 2 p$$ at parabola vertex is obtained by $$\alpha$$ rotation

$$x \sin \alpha+y \cos \alpha = p/2$$

are lines tangent to circle diameter $$p$$ not sketched for the familiar tangent to circle tangent lines not to crowd up too much in the image.

Let the variable $$\cos \alpha = u$$ for short.

$$x u -y \sqrt{1-u^2}=t,\; x \sqrt{1-u^2} + y u =\dfrac{p^2-t^2}{2p} \tag 6$$

Solve for $$(x,y)$$ as functions of $$(u,t)$$ that respectively act upon to enable tangent point change as well as rotate the entire parabola.

$$(x,y)=$$ $$\left( \dfrac {t u + (p^2 - t^2) \sqrt{1 - u^2}} {2 p}, \dfrac { u(p^2-t^2) + 2 p t \sqrt {1 - u^2}}{2 p} \right) \tag 7$$

The graph is drawn for $$p=2$$.

A non-vertex parameter $$t$$ point could be also taken for a parameter wlog. • Hi, thankyou very much for this comment its a very interesting property that I had never seen before, as the book didnt mention it. Appreciate it, it´s not quite what I need for this problem though – sara 15 Dec 5 '20 at 3:23
• Thankyou very much, sorry I cant give you a thumbs up cause i dont qualify for that yet – sara 15 Dec 5 '20 at 5:07
• Please type the Spanish original for English translation. – Narasimham Dec 5 '20 at 5:11
• ok. The original says: Se considera una parábola, cuyo parámetro p (distancia del foco a la directriz) es fijo, que gira manteniendo su foco inmóvil. Se considera también una cierta dirección fija en el plano de la parábola y se traza la tangente a ésta que tiene la dirección dada. Hallar la ecuación del lugar geométrico que describe el punto de tangencia. – sara 15 Dec 5 '20 at 9:43
• I would translate it like this: Consider a parabola whose distance $p$ from the focus to the directrix is fixed, which rotates while keeping its focus fixed. Also consider a certain fixed direction in the parabola's plane (presumably the Cartesian plane), and trace the tangent to the parabola that has that given direction. Find the equation of the locus of the points of tangency. – Toby Mak Dec 9 '20 at 7:08