Proving an ellipse fits Kepler's 1st law

[stephen cripps]

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 showr^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.

 

[mfb]

I would start with the cosine rule without the relation you want to prove, and then see if solving it for p works.
Alternatively, set up a coordinate system and work with those two coordinates and the first equation and go to r and theta later.

Independently of the approach, you can try to find $\theta_0$ first.