Solution for assignment 19 set 6.

Rewritten and using cos θ = x/r, the initial equation becomes

p = r (1 + e cos θ) = r (1 + e x/r) = r + e x or r = p − e x .

Squaring both sides

x^{2} + y^{2} = p^{2} − 2 p e x + e^{2} x^{2} .

Bringing all terms with x or y to one side, x^{2} (1 − e^{2}) + 2 p e x + y^{2} = p^{2} ,

x^{2} + \frac{2 p e}{1 − e^{2}} x + \frac{y^{2}}{1 − e^{2}} = \frac{p^{2}}{1 − e^{2}} ,

According to the recipe for completion of the square we substitute

x′ = x + \frac{p e}{1 − e^{2}}

and obtain

x′ ^{2} + \frac{y^{2}}{1 − e^{2}} = \frac{p^{2}}{1 − e^{2}} + (\frac{p e}{1 − e^{2}} )^{2} = \frac{p^{2} (1 − e^{2}) + p^{2} e^{2}}{(1 − e^{2})^{2}} = \frac{p^{2}}{(1 − e^{2})^{2}} ,

x′ ^{2} \frac{(1 − e^{2})^{2}}{p^{2}} + y^{2} \frac{1 − e^{2}}{p^{2}} = 1 .

With the definitions

a = \frac{p}{1 − e^{2}} and b = \frac{p}{\sqrt{1 − e^{2}}}

this reads

(\frac{x′}{a} )^{2} + (\frac{y}{b} )^{2} = 1 .