# Property of ellipses involving normals at the endpoints of a focal chord and the midpoint of that chord

While solving a book on ellipses, I came across the following property of an ellipse which was given without proof :-

If the normals be drawn at the extremities of a focal chord of an ellipse, a line through their point of intersection parallel to the major axis will bisect the chord.

Working with the standard ellipse, i.e., $$\frac {x^2}{a^2}+\frac {y^2}{b^2}=1,$$we find that the normal at a point $(a\cos\theta,b\sin\theta)$ is given by $$N(\theta)= a\, x \, \sec \theta- b\, y \csc \theta=a^2-b^2$$

The midpoint of $2$ parametric points $\theta$ and $\phi$ is

$$\left(\frac a2(\cos\theta+\cos\phi),\frac b2(\sin\theta+\sin\phi)\right)\equiv \left(a\cos \frac {\theta+\phi}2\cos \frac {\theta-\phi}2,b\sin \frac {\theta+\phi}2\cos \frac {\theta-\phi}2\right)$$

Since the chord is a focal chord, we have the relation that relates $\theta$ and $\phi$, i.e.,

$$\tan \frac {\theta}2\tan \frac {\phi}2=\frac {e-1}{e+1}$$

Hence the line through the midpoint parallel to the major axis is

$$y=b\sin \frac {\theta+\phi}2\cos \frac {\theta-\phi}2$$

But proceeding after this (finding the intersection of the normals at the parametric points) becomes very complicated. Is there a smarter and easier way to prove the given property except using brute force?

• yeah,the edit is ok but some answers would be appreciated – Abhishek Bakshi Feb 27 '15 at 13:47

Unlike the one I gave in this, this is a "pure geometric solution".

The figure is pretty much self-explanatory but still I will spell it out:

1. $P$ and $Q$ are points on the ellipse.
2. $PR$ and $QR$ are the tangents at these points.
3. $PC$ and $QC$ are the normals at these points.
4. Line $AB$ is a directrix.
5. $F_1$ is a focus.
6. $CD$ is drawn parallel to the major axis.

The angles of same color are equal(excuse the right angles from this rule), but for those who are color-blind or otherwise incapable of distinguishing the colors, I will spell out which angles are equal(you can figure out why):

1. $\angle PRA=\angle PCD=\alpha$
2. $\angle QRB=\angle QCD=\beta$
3. $\angle CPD=\angle PRF_1=\theta$
4. $\angle CQD=\angle QRF_1=\gamma$

Since this is an ellipse, $\dfrac{PF_1}{PA}=\dfrac{QF_1}{QB}=e\tag{i}$

Using basic trigonometry, we can write $PF_1=PR\sin\theta$,$\,PA=PR\sin\alpha$,$\,QF_1=QR\sin\gamma$ and $QB=QR\sin\beta$. Now substituting these in (i):

$$\frac{\sin\theta}{\sin\alpha}=\frac{\sin\gamma}{\sin\beta}\tag{ii}$$

Applying sine rule in $\Delta CQD$: $\dfrac{CD}{QD}=\dfrac{\sin\gamma}{\sin\beta}\tag{iii}$

Applying sine rule in $\Delta CPD$: $\dfrac{CD}{PD}=\dfrac{\sin\theta}{\sin\alpha}\tag{iv}$

Dividing (iii) and (iv), the ratio $PD:QD=\dfrac{\sin\gamma\sin\alpha}{\sin\beta\sin\theta}$

This exercise can be understood as an application of a general result about perimeter bisectors of triangles.

Proposition. Given $\triangle ABC$ with incircle $\bigcirc I$ meeting the edges at $D$, $E$, $F$ as shown. If $F^\prime$ is the point opposite $F$ in $\bigcirc I$, and if $F^{\prime\prime}$ is the point where $\overleftrightarrow{CF^\prime}$ meets $\overline{AB}$, then $$|\overline{CA}|+|\overline{AF^{\prime\prime}}| = |\overline{CB}|+|\overline{BF^{\prime\prime}}| \tag{\star}$$ so that $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$.

Proof of Proposition. Let the perpendicular to $\overline{FF^\prime}$ at $F^\prime$ meet the edges of the triangle at $A^\prime$ and $B^\prime$. By tangent properties of circles, we have $$\overline{CE}\cong\overline{CD} \qquad \overline{A^\prime E}\cong\overline{A^\prime F^\prime} \qquad \overline{B^\prime D}\cong\overline{B^\prime F^\prime}$$ Consequently, $|\overline{CA^\prime}| + |\overline{A^\prime F^\prime}| = |\overline{CB^\prime}| + |\overline{B^\prime F^\prime}|$, so that $\overline{CF}$ is a perimeter bisector of $\triangle A^\prime B^\prime C$. The Proposition holds by the similarity of $\triangle ABC$ and $\triangle A^\prime B^\prime C$. $\square$

The Proposition has a helpful corollary.

Corollary. Given $\triangle ABC$ with incenter $I$ and perimeter bisector $\overline{CF^{\prime\prime}}$, if $M$ is on $\overline{AB}$ such that $\overline{IM} \parallel \overline{CF^{\prime\prime}}$, then $M$ is the midpoint of $\overline{AB}$.

Proof of Corollary. The points of tangency of the triangle with its incircle separate the perimeter into three pairs of congruent segments, marked $a$, $b$, $c$. Thus, the semi-perimeter of $\triangle ABC$ is $a+b+c$, and since $|\overline{BC}| = b+c$, it follows that $|\overline{BF^{\prime\prime}}| = a = |\overline{AF}|$. Thus, $\overline{FF^{\prime\prime}}$ lies between congruent segments. In $\triangle FF^\prime F^{\prime\prime}$, segment $\overline{IM}$ passes through the midpoint of one side ($\overline{FF^\prime}$) and is parallel to another ($\overline{F^\prime F^{\prime\prime}}$); it necessarily meets the third side ($\overline{FF^{\prime\prime}}$) at its midpoint, which must also be the midpoint of $\overline{AB}$. $\square$

To solve the original problem, it basically suffices to embed the above triangle into an ellipse:

In the above, the ellipse's foci are $C$ and $F^{\prime\prime}$, and $\overline{AB}$ is a chord through the latter. The fundamental nature of ellipses implies that $(\star)$ holds; therefore, $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$. Moreover, the reflection property of ellipses implies that normals at $A$ and $B$ bisect angles $\angle CAF^{\prime\prime}$ and $\angle CBF^{\prime\prime}$; therefore, the intersection of these normals is the incenter of $\triangle ABC$. The result follows by the Corollary. $\square$

Referring to G-man's sketch, $\epsilon = eccentricity, RQB = \phi , F_1QR =\psi, then \frac {\cos\psi }{\cos\phi} = \epsilon$

The proof is very near from here.