# Proving an ellipse fits Kepler's 1st law

• stephen cripps
In summary, the conversation discusses the definition of an ellipse and its relation to two fixed points, known as foci. The conversation also provides a hint to solve the equation for an ellipse, involving drawing a picture and using the cosine rule. The conversation also suggests two possible approaches to solving the equation. The first approach involves using the cosine rule without the relation and then solving for p. The second approach involves setting up a coordinate system and working with the coordinates and the first equation before solving for r and theta.

## 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.

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.