The normal to an ellipse bisects the angle from its foci

Question

There is a proof on wikipedia, which proves the fact for the tangent - hence this must hold for the complementary angle. However, this was a textbook question which focused on coordinate geometry, so I do not think that it is the intended method.

The questions says:

The point $P$ lies on the ellipse with equation $9x^2 + 25y^2 = 225$, $A$ and $B$ are the points $(-4,0)$ and $(4,0)$ respectively.

(a) Prove that $PA+PB = 10$

(b) Prove also that the normal at $P$ bisects the angle $APB$.

The first part is simply showing $A$ and $B$ are the loci. The property follows by the property of loci.

I do not see how part (a) helps proving part (b) (or is it a red herring?) I tried a number of methods:

for example, let $x^*,y^*$ be a point on the ellipse. Then the normal hit the $x$ axis at C = $(16x^*/25, 0)$. Then I tried to compute the tan of angle APC and BPC by using the tangent formula. I ended up with rather ugly expressions.

Any hints?

Answer 1

Slope of normal is $$\frac{5 \sin t}{3 \cos t}$$ for $P (5 \cos t,3 \sin t)$

Eq of normal=》

$$ (y- 3 \sin t)=\frac{5 \sin t}{3 \cos t} (x- 5 \cos t)$$

$$5 x \sec t-3 y \csc t = 16$$

Hint:

Take a general point on the normal, say $(h,k)$ if perpendicular distance to line PA = perpendicular distance to line PB.

then normal is angular bisector of angle APB.

Answer 2

$PA+PB=2a$

Hence $PA/PB=2a/PB-1$

Find $PB = e(a/e- x_1)=1-ex_1$,where $x_1$ is $x$ coordinate of P. Find point of intersection of normal with $x$ axis $L$. Find $AL/BL$. Use of parametric coordinates simplifies calculations.

Padmanabhan

Answer 3

Here is a rather nice differential argument in a general setting that works for all conics as we are going to show it.

An ellipse can be defined in different equivalent ways. Among the most important are these two ones:

• as the set of points $M(x,y)$ such that

$$\tag{1}MF+MF'= k.$$

where $k$ is a constant, and $F$ and $F'$ are the foci (have a look at the first figure below). We are going to assume that the axes are chosen in such a way that the coordinates of $F$ and $F'$ are resp. $(f,0)$ and $(-f,0)$.

• as a parameterized curve, the most simple being, with respect to the axes just described:

$$\tag{2}\begin{cases}x&=&a \cos(t)\\y&=&b \sin(t)\end{cases}$$

(with $a^2=b^2+f^2$).

Analytically, (1) becomes:

$$\tag{3}f(x,y)=\sqrt{(x-f)^2+y^2}+\sqrt{(x+f)^2+y^2}=k$$

If we consider $x$ and $y$ defined by (2), $f$ is a function of $t$ through $x$ and $y$ ; let us differentiate (3) with respect to $t$

$$\dfrac{\partial f}{\partial t}=0 \ \ \ \iff \ \ \ \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial t}=0$$

which is equivalent, using the fact that $\sqrt{u}$ is differentiated as $\tfrac{u'}{2 \sqrt{u}}$, to:

$$\left(\tfrac{(x-f)}{\sqrt{(x-f)^2+y^2}}+\tfrac{(x+f)}{\sqrt{(x+f)^2+y^2}}\right)\dfrac{\partial x}{\partial t}+\left(\tfrac{y}{\sqrt{(x-f)^2+y^2}}+\tfrac{y}{\sqrt{(x+f)^2+y^2}}\right)\dfrac{\partial y}{\partial t}=0.$$

Let us rearrange the expression above in the following way:

$$\tag{4} \dfrac{1}{\sqrt{(x-f)^2+y^2}}\left((x-f)\dfrac{\partial x}{\partial t}+y\dfrac{\partial y}{\partial t}\right)+ \dfrac{1}{\sqrt{(x+f)^2+y^2}}\left((x+f)\dfrac{\partial x}{\partial t}+y\dfrac{\partial y}{\partial t}\right)$$

In (4), we recognize dot products with $\vec{T}$, the tangent vector in $M$. This gives:

$$\tag{5}\dfrac{1}{\|\vec{FM}\|}\vec{FM} . \vec{T}+\dfrac{1}{\|\vec{F'M}\|}\vec{F'M} . \vec{T}=0 \ \ \iff$$

$$\tag{6}\underbrace{\left(\dfrac{1}{\|\vec{FM}\|}\vec{FM}+\dfrac{1}{\|\vec{F'M}\|}\vec{F'M}\right)}_{\text{directing vector of the angle biss. (*)}} . \vec{T}=0$$

expressing that the directing vector of the angle bissector of $\widehat{FMF'}$ is orthogonal to the tangent vector, thus is a directing vector of the normal line to the ellipse at $M$, ending the proof.

Remarks:

1) One can be convinced of characterization $(*)$ by thinking to the "manual" construction of an angle bissector using a compass with the same aperture on $MF$ and $MF'$.

2) This method works in the same way for parabolas considered as limit cases of ellipses for which foci $F'$ tends to infinity.

3) A different phenomena occurs for hyperbolas. One should consider one branch at a time (see fig. 2 below). For example, the branch for which (see the difference with (1)

$$\tag{7}MF\color{red}{-}MF'=k$$

It is enough to change all along the calculations above, the central "plus" sign into a minus sign, and at the end obtain:

$$\tag{8}\underbrace{\left(\dfrac{1}{\|\vec{FM}\|}\vec{FM}\color{red}{-}\dfrac{1}{\|\vec{F'M}\|}\vec{F'M}\right)}_{\text{directing vector of the angle biss. (*)}} . \vec{T}=0$$

This minus sign explains the fact that the rays issued from $F$ (red point), reflect on the hyperbola and, instead of passing through the second focus $F'$, make it a virtual focus, in the sense that, for an observer situated on the right, looking leftwads, the rays seem to emanate from $F'$.

4) This method is connected with the "first variation formula" (see p. 104 of "Problems in Geometry", Marcel Berger, P. Pansu, J.-P. Berry, X. Saint-Raymond, Springer.)

Fig. 1: Illustration of formula (6). In red, the unit vectors $\tfrac{1}{\|\vec{FM}\|}\vec{FM}$ and $\tfrac{1}{\|\vec{F'M}\|}\vec{F'M}$; the sum of these vectors is a directing vector of the normal.

Fig. 2: Illustration of formula (8).

Remark : take a look page 12 of the interesting document (https://www.math.psu.edu/tabachni/Books/billiardsgeometry.pdf)

Answer 4

This can be shown using a geometric argument if we look at an infinitesimal arc, $\overset\frown{AB}$ of the ellipse.

An infinitesimal arc is essentially a line, so we will consider $\overline{AB}$. Furthermore, we essentially have $$ \overline{F_1A}\parallel\overline{F_1B}\quad\text{and}\quad\overline{F_2A}\parallel\overline{F_2B} $$ Also, let $$ \overline{AD}\perp\overline{DB}\quad\text{and}\quad\overline{BC}\perp\overline{CA} $$ Due to the property that the sum of the distances from the two foci to any point on the ellipse is constant, we also have, essentially, $$ \overline{DB}=\overline{CA} $$ Therefore, being right triangles with equal hypotenuses and legs, $$ \triangle ADB\simeq\triangle BCA $$ This means that $$ \angle ABD=\angle BAC $$ This means that the perpendicular to $\overline{AB}$ bisects $\angle F_1AF_2$.