Deriving polar coordinate form of ellipse. Issue with length of a distance to a foci.

Question

I am reading through in Spivak on how to obtain the polar coordinate form of the ellipse. I'm given the following diagram:

All I'm trying to do is establish that the distance between $(x,y)$ and $(-2\epsilon a, 0)$ is as is stated, i.e: $2a - r$

I didn't think this would be much of an issue. I actually attempted to apply the usual trigonometric ideas on right triangles getting the expression:

$$(x - (-2 \epsilon a))^{2} + y^{2}$$

expand things out and take a square root and voila I would be done.....But that is not actually happening. To get rid of $\epsilon$ I could just assume it to be $1$, but the algebra when I work it out leaves me stuck with the following:

$$4a^{2} + 4xa + x^{2} + y^{2}$$

I can see where the $r$ term will appear, but it won't be a negative. As well I don't see how it will simplify to $2a$.

I did try to reverse engineer things by using the expression $(2a - r)^{2}$ to try and derive the original expression, but what I got from that derivation was:

$$(2a - r)^{2} = 4a^{2} - 4ar + r^{2}$$

where $r^{2} = x^{2} + y^{2}$

Messing around with the first and second expressions and subtracting them from one another I am left with notion that $x = -r$

But based on everything I did in the first expression I don't see how that can come about......SOme help would be appreciated.

Answer 1

In an ellipse the sum of the distances from the foci is constant.

Traditionally we say that this distance is $2a.$ And $2a$ is also the length of the major axis.

The distance from focus to $(x,y) = (2a-r)$ is a direct consequence of the definition.

$\epsilon$ is not going to cancel out in the algebra. $\epsilon$ defines the shape of the ellipse, and $a$ defines the size. As $\epsilon$ approaches $1$ the ellipse becomes so stretched out that it cannot actually close and is, in fact, a parabola.

As $\epsilon$ approaches $0$ the foci move right next to each other and the figure becomes a circle.

Where do you go from here?

$(x+2\epsilon a)^2 + y^2= (2a - r)^2\\ x^2 + y^2 + 4x\epsilon a = 4a^2 - 4ar + r^2\\ x^2 + y^2 = r^2\\ x\epsilon = a - r\\ x = r\cos \theta\\ r(\epsilon\cos\theta + 1) = a\\ r = \frac {a}{\epsilon\cos\theta + 1}$

Answer 2

In Spivak's derivation two things are Given:

1) sum of two distances $= 2a,$ a given property

2) distance to focus $d_{f-P}$

Required to come to Newton's polar form.

That is straight forward algebraic simplification.

LHS

$$ (2a-r)^2 = 4 a^2 +r^2 -4 a r = 4 a^2 --4 a r +x^2 +y^2 $$ RHS $$ x^2 + 4 \epsilon^2 a^2 - 4 \epsilon a x +y^2 $$

Cancel $(x^2+y^2)$ on either side

$$ a(1-\epsilon ^2)-r = -\epsilon x\quad p-r= - \epsilon x$$

or since $ x=r \cos \theta$ and first term is semi latus-rectum $p$, it results in polar form:

$$ \dfrac{p}{r}= 1- \epsilon \cos \theta $$