From standard Newtonian form for focal conics $ p/r = ( 1- \epsilon \cos \theta), $ I obtained by differentiating with respect to arc:

$$ \dfrac{FN}{p} = \dfrac{\cos \phi}{\sin \theta}. $$

enter image description here

where $ FN $= perpendicular length dropped from focus F on tangent at P and, $p$ is the semi-Latus Rectum. Please suggest a geometric construction for its proof.


I assumed drawing new construction lines would make one see a geometrical relationship.

I am unable to immediately think out a line of common shared length $$ FN \sin \theta = p \cos \phi $$

or a line of length $$ FN /\cos \phi = p /sin\theta $$ or any other.


$ (p \cos \phi) $ does not change even if we consider the other focus.It may relate to reflection angle between foci at NPF to the tangent.


The following construction I have completed in continuation with Lee David Chung Lin's setting as an aid. Take the variable point $T$ as the generating circle diameter. Draw a circle with $F$ as center and $p$ as radius. Angle $\theta $ is both an alternate angle between two parallel lines as well as it is contained in the alternate segment making $ FST, FBR $ triangles congruent. $TS$ is perpendicular to ellipse radius vector $FP$ and tangent to circle of radius $p$ at $S$.

$$ FS= p= FT \,\sin \theta $$

$$ FN= FT \,\cos \phi $$


$$ \dfrac{FN}{p} = \dfrac{\cos \phi}{\sin \theta} $$


  • $\begingroup$ So you're looking for a geometric proof of the fact that $\dfrac{FN}{p} = \dfrac{\cos \phi}{\sin \theta}$ , right? $\endgroup$ – G-man Apr 19 '15 at 9:09
  • $\begingroup$ Won't $p$ be the semi-latus rectum? $\endgroup$ – G-man Apr 19 '15 at 14:41
  • $\begingroup$ Nicely done! This is exactly what a construction should look like! $\endgroup$ – Lee David Chung Lin Sep 29 '16 at 0:11

As shown in the figure just below, there's a special relation between the angles $\theta$ and $\phi$ that gives the desired equation for lengths:

Point $P$ is on the ellipse. Its tangent makes angle $\phi$ with the major axis (taken to be the horizontal $x$-axis WOLG).

Make the respective line from the two foci $F_1$ and $F_2$ perpendicular to the tangent. The intersections with the tangent are $N_1$ and $N_2$, and the intersections with the horizontal line (that passes through $P$) are $Q_1$ and $Q_2$, respectively.

Since $P$ is on the ellipse, $~\overline{F_1 P} = r~$ makes $~\overline{F_2 P} = 2a-r ~$, where $a$ is the semi-major axis.

The angle of incidence equals the angle of reflection $~\angle F_1 P N_1 = \theta - \phi = \angle F_2 P N_2~$, we have $$\begin{align} \overline{N_1N_2} &= \overline{N_1P} + \overline{N_2P} \\ &= \overline{F_1P}\cos(\theta - \phi) + \overline{F_2P}\cos(\theta - \phi) \\ &= 2a \cos(\theta - \phi) \end{align}$$

The length of the horizontal line segment $\overline{Q_1 Q_2}$ is obviously the same as the distance between the two foci, that is $$\overline{N_1N_2} \frac{1}{ \cos\phi } = \overline{Q_1Q_2} = \overline{ F_1 F_2} = 2\sqrt{a^2 - b^2}= 2c = 2ae$$ where $e < 1$ is the eccentricity of the ellipse.

Hence, we arrive at this nice relation for any point $P$ on the ellipse that $$\begin{align} \frac{\cos(\theta -\phi)}{\cos\phi} &= e &&\text{or} &\left|\frac{\hat{\dot{r}} \cdot \hat{r} }{ \hat{\dot{r}} \cdot \hat{s} } \right| &= e \end{align}$$ where $\hat{r}$ the unit position vector of an orbiting object, $\hat{\dot{r}}$ is its unit velocity (tangent) vector, and $\hat{s}$ is the unit vector along the major axis. Clearly this invariance can be stated in different forms (e.g. with $\vec{r}$ and not $\hat{r}$, with momentum and not $\hat{\dot{r}}$) that might be useful in various physical analysis.

The algebraic derivation of this angle relation is rather tedious, but this invariance along the orbit is so elegant and powerful, bringing many nice geometric results, I wonder why this is not prominently showcased in the physics textbooks (celestial mechanics or just classical mechanics) I've read$\ldots$ or maybe it's just been too long that I don't recognize it anymore.

Now, I'm going to establish a right triangle using the angle relation. Observe in the figure below that $$\overline{CP} = \overline{QP} \frac{1}{ \cos\phi } = \frac{r}{ e\cos\phi }$$ since point $P$ being on the ellipse means $~\overline{PF} = e\cdot d(P, \Gamma) = e\cdot \overline{QP}~$, where $~d(P, \Gamma)~$ is the distance between point $P$ and the directrix $\Gamma$.

enter image description here

At the same time, $~\angle CPF = \angle QPF - \angle QPC = \theta - \phi~$ therefore $$\begin{align} \overline{CP} \cdot \cos(\angle CPF) &= \overline{CP} \cdot \cos(\theta - \phi) \\ &= \overline{CP} \cdot e\cos(\phi) \\ &= \frac{r}{ e\cos\phi } \cdot e\cos(\phi) = r = \overline{FP} \end{align}$$

Namely, this proofs that it is a right angle at $\angle CFP$, and we have the angles as labeled in the figure.

We can proceed to the relation involving the semi-latus rectum $~p=\overline{FB}~$:

$$\overline{FB} \cdot \frac{\cos\phi}{ \sin\theta } = \frac{\overline{FB}} e \frac{\cos(\theta - \phi) }{ \sin\theta } = \overline{AB} \frac{\cos(\theta - \phi) }{ \sin\theta }$$

since again for point $B$ on the ellipse $~\overline{BF} = e\cdot d(B, \Gamma) = e\cdot \overline{AB}~$.

With the angle $\angle CFB = \theta$ we continue with the ratio $$ p \cdot \frac{\cos\phi}{ \sin\theta } = \frac{\overline{AB} }{ \sin\theta } \cdot \cos(\theta - \phi) = \overline{CF} \cdot \cos(\theta - \phi) = \overline{FN} \tag*{**Q.E.D**} $$

  • $\begingroup$ Your analysis by equations is very nice,correct and extremely lucid. But I was n't looking for a geometric/trig proof with lengths expressed in terms variables but for a construction that makes equality of projected line segments and so on obvious. As you say I also wondered why it is not more prominently showcased in physics and celestial mechanics and in fact is one reason for my this post you first answered after a gap of 18 months. $\endgroup$ – Narasimham Sep 28 '16 at 9:33
  • $\begingroup$ Yeah I think this inquiry of yours is one of the unfortunate posts that got overlooked. I worked on it because I share your views. I also wanted a construction that `brings' one length to another, but as of now it still eludes me and the above is what I got so far. I might continue to work on it$\ldots$ I don't know, maybe I need to start fresh and think in terms of vector analysis instead of plane geometry. $\endgroup$ – Lee David Chung Lin Sep 28 '16 at 10:01
  • $\begingroup$ btw, perhaps not a direction you're interested in, but I think this might be related to how one can solve problems in mechanics like using Snell's Law in the brachistchrone problem. The un-constrained striaghtline (under uniform gravity) and conics (under inverse-square gravity) are all special solutions of brachistochrone that run their "natural courses". I made this connection because Snell's law is about the angle made between the tangent vector of the trajectory and the normal vector of the potential field. $\endgroup$ – Lee David Chung Lin Sep 28 '16 at 10:18
  • $\begingroup$ I for one would love to see the principles of optics and mechanics treated in the same framework. Even after four centuries, the topic still holds fundamental basic STEM interest. E.g., one link is tf.uni-kiel.de/matwis/amat/admat_en/kap_5/advanced/t5_1_1.html ; Maybe you can ask to show in a different question a unifying derivation (light/mechanics) that would interest many here. I had attempted it with limited success. As for this post, I shall soon upload a sort of relevant construction step. $\endgroup$ – Narasimham Sep 28 '16 at 15:53
  • $\begingroup$ Looking forward to it. So far I have a new proof that is still just halfway there. I have to start with knowing $\angle CFP$ is a right angle, then it can be shown that the desired equation is equivalent to $~\overline{CF}\cdot \overline{FP} = \overline{FM}\cdot \overline{PQ}~$ where point $M$ is the intersection of the $y$-axis with $\overline{PN}$. The equation holds since the area of a right triangle is the same as the rectangle with that height $\overline{FM}$. $\endgroup$ – Lee David Chung Lin Sep 28 '16 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.