Ellipse in Polar Coordinates
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A short note by Phil dated 11.13.16. It shifts a centered ellipse horizontally by c, substitutes x = r cosθ, y = r sinθ, b and c = εa, and solves the resulting quadratic in r with Maple. It checks the two roots r± at θ = π and θ = 0, shows only one is valid for each shift direction, and notes θ spans 0 to 2π since the focus lies inside. Equations and figures were lost in extraction.
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Ellipse in Polar Coordinates PhL 11.13.16
Start with an ellipse centered at the origin,
+ = 1
We want to shift this horizontally to the right so that the center of the ellipse is at xc = +c.
+ = 1 (1)
which says the xc = c. Here is a picture of this shifted ellipse
This puts the left focal point at the origin!
Write as (1) as :
f ≡ b2(x-c)2 + y2a2 - a2b2 = 0 .
Now make these replacements:
x = rcosθ
y = rsinθ
b =
c = εa
and do it all in Maple:
There are two solutions which are
r± =
Why are there two solutions? Well, it was a quadratic equation in r. I know from my setup above if we set θ = π we should get r = a-c (see picture) . Let's test the two results
r+ = = = a(1+ε) = a + c
r- = = = a(1-ε) = a -c
Only r- is correct, while r+ is spurious for my positioning of the ellipse.
Suppose I started shifting the other way. That is like taking c → -c and that is like ε → -ε. So in that case the correct solution is r+.
Here are the conclusions:
Fact: In either polar coordinate representation of the ellipse, the origin is at a focal point of the ellipse.
Final check:
r+(θ=0) = = a(1-ε) = a - c correct
r-(θ=0) = = a(1+ε) = a + c correct
Just as a final look, here are the angles θ for the two ellipse cases
In each case θ takes the full range 0 to 2π since the origin lies inside the ellipse.