# Proving an ellipse fits Kepler's 1st law

## Homework Statement

(This isn't coursework, just a revision question)
Exercise: An ellipse can be defined as the locus of all points, P, in the plane such that
$$PF_1+PF_2=\frac{2p}{(1-ε^2)}$$
where F1 and F2 are two fixed points, and PF1 is the distance from P to F1 (similarly, P F2).
F1 and F2 are known as the foci. By placing F1 at the origin, and F2 at $$x = \frac{-2pε}{(1-ε^2)}$$
show that the ellipse satisfies $$r = \frac{p}{(1+εcos(θ- θ_0))}$$

## Homework Equations

The example is on page 26 of this pdf http://www.nottingham.ac.uk/~ppzap4/PoD.pdf
Hint: Draw a picture and use the cos rule

## The Attempt at a Solution

I've assumed that PF1 is r, and therefore PF2 is $$\frac{2p}{(1-ε^2)}-r$$ and have used the cosine rule to show$$r^2 = (\frac{p}{(1+εcos(θ- θ_0))})^2=(\frac{2p}{(1-ε^2)}-r)^2+(\frac{-2pε}{(1-ε^2)})^2+(\frac{2p}{(1-ε^2)}-r)(\frac{-2pε}{(1-ε^2)})cos(θ- θ_0)$$I've tried multiple times but can't seem to rearrange this to prove it is true, can anyone give me a pointer in the right direction.