Showing that certain points lie on an ellipse

Question

I have the equation $$r(\phi) = \frac{es}{1-e \cos{\phi}}$$ with $e,s>0$, $e<1$ and want to show that the points $$ \begin{pmatrix}x(\phi)\\y(\phi)\end{pmatrix} = \begin{pmatrix}r(\phi)\cos{\phi}\\r(\phi)\sin{\phi}\end{pmatrix}$$ lie on an ellipse.

I have a basic and a specific problem which may be related. The basic problem is that I don't see how to get to the goal.

I want to arrive at an ellipse equation $(x/a)^2 + (y/b)^2 = 1$ and am confused that we already have

$$\left(\frac{x(\phi)}{r(\phi))}\right)^2 + \left(\frac{y(\phi)}{r(\phi))}\right)^2 = 1$$

This can't be the right ellipse equation because it has the form of a circle equation and obviously, we don't have a circle here. But how should the equation I aim for look like then?

I started calculating anyway and arrived at a specific problem. Substituting $x(\phi)$ into $r(\phi)$, I got $r(\phi)=e(s+x(\phi))$ and using $r(\phi)^2 = x(\phi)^2 + y(\phi)^2$ I got an equation which doesn't involve $r(\phi)$ anymore but has $x(\phi)$ in addition to $x(\phi)^2, y(\phi)^2$.

If I could eliminate the linear term somehow, the resulting equation is an ellipse equation. But I don't see a way how to accomplish this.

Answer 1

Hints: Your ellipse is not centered at the origin. Substitute $\phi=0$ and $\phi=\pi$ to get the ends of the major axis. You get $\left(-\frac{es}{1+e},0\right)$ and $\left(\frac{es}{1-e},0\right)$ as the left and right axis ends.

The center of the ellipse is halfway between those two points.

The standard equation of an ellipse centered at point $(h,k)$ is

$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$

The semi-major axis value $a$ is half the distance between the two ends of the major axis.

So find point $(h,k)$ and value $a$. Then find $b$ and show the equation works.

Note: The linear term in $x$ should not be "removed" but is involved in completing the square so you get $(x-h)^2$.

Answer 2

Your work seems correct. I suppose that from $x^2+y^2=e^2(s+x)^2$ you find: $x^2(1-e^2)+y^2-2se^2x-e^2s^2=0$, and this is an equation of the form: $$ Ax^2+Bxy+Cy^2+Dx+Ey+F=0 $$ that represents a conic section, and it is an ellipse if $B^2-4AC<0$, as it is easely verified since $0<e<1$.

Answer 3

When $ r$ and $ x $ are together present for ellipse with origin at one focal point,

$$ es = r - e \,x $$

I mention this on purpose even if it adds to the confusion.. that you would eventually resolve after knowing difference between central and focal ellipses.

Answer 4

I'm guessing that the equation you got that "doesn't involve $r(\phi)$ anymore" is equivalent to this:

$$(1-e^2)(x(\phi))^2 - (2e^2s) x(\phi) + y(\phi)^2 = e^2s^2. \tag 1$$

I'm going to "cheat" a little here: there happens to be a well-known formula,

$$r = \frac{a(1-e^2)}{1 - e \cos{\theta}}, \tag 2$$

for the polar coordinates $(r,\theta)$ of an ellipse with semimajor axis $a$ and eccentricity $e$, with one focus at the origin. This gives me the idea that I should set a parameter $a = \frac{es}{1-e^2},$ because then I will have $es = a(1-e^2).$ The original equation then becomes Equation $(2)$, and after substituting for $es$, Equation $(1)$ becomes

$$(1-e^2)(x(\phi))^2 - (2a(1-e^2)e)\, x(\phi) + (y(\phi))^2 = a^2(1-e^2)^2. $$

In preparation for the technique of completing the square, it's helpful for the leading coefficient of the interesting squared variable ($(x(\phi))^2$ in this question) to be a square. Typically we would want it to be $1$, but in this case I have a strong hunch we will want to see something like $\left(\frac{x - x_0}{a}\right)^2$ in the final equation, so I'll try $\frac{1}{a^2}$ as the leading coefficient. We get there by dividing both sides by $a^2(1-e^2)$:

$$\frac{x(\phi)^2}{a^2} - 2e\frac{x(\phi)}{a} + \frac{(y(\phi))^2}{a^2(1-e^2)} = 1 - e^2. $$

Now to complete the square, we use the fact that $(u+v)^2 = u^2 + 2uv + v^2$ for any $u$ and $v$; let $u^2$ be $\frac{x(\phi)^2}{a^2}$ and $2uv$ be $-2e\frac{x(\phi)}{a}$, that is, $u = \frac{x(\phi)}{a}$ and $v = -e$, so

$$\left(\frac{x(\phi)}{a} - e\right)^2 = \frac{(x(\phi))^2}{a^2} - 2e\frac{x(\phi)}{a} + e^2.$$

So we can substitute $\left(\frac{x(\phi)}{a} - e\right)^2 - e^2$ for $\frac{(x(\phi))^2}{a^2} - 2e\frac{x(\phi)}{a}.$ Do so:

$$\left(\frac{x(\phi)}{a} - e\right)^2 - e^2 + \frac{(y(\phi))^2}{a^2(1-e^2)} = 1 - e^2. $$

We can nicely cancel the constant $e^2$ term on each side:

$$\left(\frac{x(\phi)}{a} - e\right)^2 + \frac{(y(\phi))^2}{a^2(1-e^2)} = 1. $$

It should be reasonably clear already that this is an ellipse, but if we set $b = a\sqrt{1-e^2}$ and $x_0 = ae,$ we can massage this equation into the form $$ \left(\frac{x(\phi) - x_0}{a}\right)^2 + \left(\frac{y(\phi)}{b}\right)^2 = 1, $$ which is the equation for an ellipse whose semi-major axis (parallel to the $x$ axis) is $a$, whose semi-minor axis is $b$, and whose center is at $(x_0,0)$.

Answer 5

Let us rewrite $$r(\phi) = \frac{es}{1-e \cos{\phi}}$$ in Cartesian coordinates $$r(\phi)(1-e \cos{\phi}) =r(\phi)-ex(\phi)= es,$$ then, $$r^2(\phi)-(ex(\phi)+es)^2=(1-e^2)x^2(\phi)+y^2(\phi)-2e^2sx(\phi)-e^2s^2=0.$$

By completing the square, we rewrite as $$(1-e^2)\left(x(\phi)-\frac{e^2s}{1-e^2}\right)^2+y^2(\phi)-\frac{e^2s^2}{1-e^2}=0,$$ or $$\frac{\left(x(\phi)-\frac{e^2s}{1-e^2}\right)^2}{\frac{e^2s^2}{(1-e^2)^2}}+\frac{y^2(\phi)}{\frac{e^2s^2}{1-e^2}}=1,$$

an ellipse centered at $(\dfrac{e^2s}{1-e^2},0)$ with half-axes $\dfrac{es}{1-e^2}$ and $\dfrac{es}{\sqrt{1-e^2}}$.