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Let $P \in RP^2$ fixed. Let C be a conic. Consider all lines $L_{AB}$ such that there exist $A,B \in C$ such that $\measuredangle APB = \frac{\pi}{2}$. Prove that the set of lines $L_{AB}$ envelope a curve of class 2.

I think Laguerre's Formula can be used to prove this, and I need help for some strategy.

roger98
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  • You mean I should work on the dual of $RP^2$? – roger98 Mar 09 '17 at 06:56
  • No, there is a misunderstanding: you say that a set of lines is a curve. A set of lines usually is not a curve : at most, it envelopes a curve. It becomes a curve in the dual. – Jean Marie Mar 09 '17 at 07:00
  • Ok. I changed the description of the question. – roger98 Mar 09 '17 at 07:08
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    Thanks for your advice. I have edited it again. – roger98 Mar 09 '17 at 07:42
  • If you'd like to attract more attention, you might also consider making the title more specific. I can't speak for everyone but I skip over dozens of questions with useless titles like this one simply because I see more specific titles with question that I know I can answer. Basically there is no excuse to write something like "A [tag-on-your-question] question". That is no more useful than the list of tags. – rschwieb Mar 10 '17 at 21:22
  • Thanks. I have added something in my title. – roger98 Mar 11 '17 at 06:20

1 Answers1

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Here is a sketch of resolution that uses a projective method at the end.

  • 1) Take point $M$ as the origin of coordinates.

  • 2) Describe your conical curve with polar coordinates $p=p(\theta)$. This is of course especialy simple when $M$ is a focus of the conic (in this case $p=\dfrac{p_0}{1+e \cos(\theta-\theta_0)}$).

  • 3) Write equation of $L_{AB}$ under the following determinant form (which, btw, is an alignment constraint of points $M,A,B$, interpreted in projective space !):

$$\begin{vmatrix}x&p(\theta)\cos(\theta)&p(\theta+\pi/2)\cos(\theta+\pi/2)\\ y&p(\theta)\sin(\theta)&p(\theta+\pi/2)\sin(\theta+\pi/2)\\ 1&1&1\end{vmatrix}=0 \ \ \iff$$

$$\begin{vmatrix}x&p(\theta)\cos(\theta)&-q(\theta)\sin(\theta)\\ y&p(\theta)\sin(\theta)&q(\theta)\cos(\theta)\\ 1&1&1\end{vmatrix}=0 \ \ \ \text{with} \ q(\theta):=p(\theta+\pi/2)\ \ \iff$$

$$\tag{1}x\underbrace{(p(\theta)\sin(\theta)-q(\theta)\cos(\theta))}_{r(\theta)}-y\underbrace{(p(\theta)\cos(\theta)+q(\theta)\sin(\theta))}_{s(\theta)}+\underbrace{p(\theta)q(\theta)}_{t(\theta)}=0$$

  • 4) Form the equation obtained by differentiating (1) with respect to $\theta$:

$$\tag{2}x r'(\theta)-y s'(\theta)+t'(\theta)=0$$

I haven't enough time to treat the general case.

Nevertheless, I have validated the method in a particular case, with the following parabola with focus (point $M$) at the origin, represented by polar curve:

$$p(\theta):=\dfrac{1}{1+ \cos(\theta)}$$

(see graphics below, generated by a Matlab program: parabola in red, lines $L_{AB}$ in blue, envelope curve in green).

enter image description here

One easily finds that the equation of $L_{AB}$ is:

$$x(\sin(\theta)-\cos(\theta)-1)+y(-\sin(\theta)-\cos(\theta))+1=0$$

The equation obtained by differentiation with respect to $\theta$ is:

$$x(\cos(\theta)+\sin(\theta))+y(-\cos(\theta)+\sin(\theta))=0$$

Solving the system formed by these two equations gives the parametric equations of the envelope:

$$\tag{3}X=\dfrac{\cos(\theta)-\sin(\theta)}{2+\cos(\theta)-\sin(\theta)}, \ \ \ Y=\dfrac{\cos(\theta)+\sin(\theta)}{2+\cos(\theta)-\sin(\theta)}$$

Important remark: (3) can be expressed as the image of unit circle $(x=\cos(\theta),y=\sin(\theta))$ by projective transformation

$$\tag{3}X=\dfrac{x-y}{x-y+2}, \ \ \ Y=\dfrac{x+y}{x-y+2}$$

thus is a conic. We recognize here a full ellipse.

Finding the locus as the image of unit circle by a projective transformation should be the case in a more general context (to be checked).

(the ellipse here has semi-major axis $\sqrt{2}$ and semi-minor axis $1$.)

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Jean Marie
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  • Thanks for your answer. It works but it's hard to calculate by hand. I want to know how to prove it by pure projective geometry. – roger98 Mar 10 '17 at 04:51