Equation general of conic in polar coordinates

AI Thread Summary
The discussion focuses on the conic equation in polar coordinates, highlighting the known parametric form, r = p/(1 + e cos(θ). Participants express curiosity about a general form of the conic equation in polar coordinates, noting that while it can be derived from the Cartesian version, it may appear complex. The conversation emphasizes that the parametric form is more suitable for polar coordinates, especially when the conic's center holds significance. There is a debate on whether seeking a general expression independent of Cartesian coordinates is necessary, with some suggesting that transformation is a natural approach. Ultimately, the complexity of deriving a general polar form is acknowledged, leaving the topic open for further exploration.
Jhenrique
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The conic equation has 2 versions in cartesian coordinates:

The general: ##Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0##
And the parametric: ##y^2 = 2px + (e^2-1)x^2##

In polar coordinates, I known just the parametric: ##r = \frac{p}{1+e\cos(\theta)}##
But exist a general form too?
 
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You can express x and y in your general equation in terms of r and θ to get the general form. I think it will look quite messy. The parametric uses the "right" coordinate system, where the equation has a nice form. Usually there is no point in polar coordinates if the center does not have a special meaning.
 
mfb said:
You can express x and y in your general equation in terms of r and θ to get the general form. I think it will look quite messy. The parametric uses the "right" coordinate system, where the equation has a nice form. Usually there is no point in polar coordinates if the center does not have a special meaning.

But what you are suggesting is a transformation... I'm not looking for this, but yes by a general expression/format that is a conic but is independent of the cartesian format...
 
Well, you look for the result of a transformation. Isn't it natural to just do this transformation then?
 
Ahhh, does not matter... thanks of anyway
 
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