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## 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

[tex]PF_1+PF_2=\frac{2p}{(1-ε^2)}[/tex]

where F

_{1}and F

_{2}are two fixed points, and PF

_{1}is the distance from P to F

_{1}(similarly, P F

_{2}).

F

_{1}and F

_{2 }are known as the foci. By placing F

_{1}at the origin, and F

_{2}at [tex]x = \frac{-2pε}{(1-ε^2)}[/tex]

show that the ellipse satisfies [tex]r = \frac{p}{(1+εcos(θ- θ_0))}[/tex]

## 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 PF

_{1}is r, and therefore PF

_{2}is [tex]\frac{2p}{(1-ε^2)}-r[/tex] and have used the cosine rule to show[tex]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)[/tex]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.