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    I am interested in how the following formula for conics is derived;

    r = \frac{b^2}{a(1+e cos\theta)}

    Which gives:

    r = \frac{a(1-e^2)}{1+ecos\theta} for an ellipse, and

    r = \frac{a(e^2 - 1)}{1+ecos\theta} for a hyperbola

    I am able to derive the two individual formulas for the ellipse and hyperbola by geometric consideration but have trouble with the one containing the term in b. Any help is appreciated .

    EDIT:

    I forgot to mention that the focus is at (ae, 0) and the directrix is x = a/e
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    Do these substitutions into the Cartesian equations not work?

    

x = r cos(\theta)

    

y = r sin(\theta)
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    (Original post by Qwertish)
    Do these substitutions into the Cartesian equations not work?

    

x = r cos(\theta)

    

y = r sin(\theta)
    The thing is that they have different Cartesian equations. I know how to derive the two formulas for ellipse and hyperbola individually. It is the first formula I mentioned that I have trouble with.
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    (Original post by Ateo)
    I am interested in how the following formula for conics is derived;

    r = \frac{b^2}{a(1+e cos\theta)}

    Which gives:

    r = \frac{a(1-e^2)}{1+ecos\theta} for an ellipse, and

    r = \frac{a(e^2 - 1)}{1+ecos\theta} for a hyperbola

    I am able to derive the two individual formulas for the ellipse and hyperbola by geometric consideration but have trouble with the one containing the term in b. Any help is appreciated .

    EDIT:

    I forgot to mention that the focus is at (ae, 0) and the directrix is x = a/e
    Right, let's do this in polar coordinates.

    Put the pole at the focus and have the initial line be perpendicular to the directrix at x = \frac{a}{e}.

    The distance from the focus to a general point on the curve is r, which depends on \theta.

    The distance from the directrix to the point will then be \frac{a}{e} - rcos\theta - ae.

    Since we are dealing with a conic section, by definition we have
    r = e(\frac{a}{e} - rcos\theta - ae)

    Which then gives
    r(1 + ecos\theta) = a(1 - e^2)

    Now 1 - e^2 = \frac{b^2}{a^2} and thus the result follows:
    r = \frac{b^2}{a(1 + ecos\theta)}
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    (Original post by Brister)
    ...
    So you've basically considered the ellipse and derived the formula for this case. Then you used that formula to get the general equation. You can also consider the hyperbola in a similar way getting the other equation I mentioned which then you can use to get the general formula mentioned.

    So essentially what we are doing is considering two cases from which the general formula follows. I already understand how to do this. What I was curious about is how the general formula is obtained without this individual consideration of each case. My book states this general formula and then gives the two separate cases. We have went in reverse. With that said though, I don't know if there is a direct proof of this, the book isn't great at introduction of theorems - it just seems to state them. Sorry for being a pain.
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    (Original post by Ateo)
    So you've basically considered the ellipse and derived the formula for this case. Then you used that formula to get the general equation. You can also consider the hyperbola in a similar way getting the other equation I mentioned which then you can use to get the general formula mentioned.

    So essentially what we are doing is considering two cases from which the general formula follows. I already understand how to do this. What I was curious about is how the general formula is obtained without this individual consideration of each case. My book states this general formula and then gives the two separate cases. We have went in reverse. With that said though, I don't know if there is a direct proof of this, the book isn't great at introduction of theorems - it just seems to state them. Sorry for being a pain.
    Okay, I see what you are trying to say. I believe, though, that this general equation involving b^2 is just a convenient way of combining the two equations you get for the ellipse and hyperbola.

    You will notice that b^2 is defined differently for the ellipse and hyperbola. This is because the value of a^2(1 - e^2) is negative for a hyperbola (e > 1) so they decide that b^2 = a^2(e^2 - 1) instead.

    In other words, your general equation cannot define an ellipse and a hyperbola at the same time.

    You cannot really derive the general equation without specifying the type of conic because the focus and directrix move depending on e. For a hyperbola, you will find that the directrix moves to the left of the curve whereas for the ellipse it is on the right.

    Circle
    It is interesting to note that the equation works for a circle, which is basically a special case of the ellipse.
    With e = 0, we first notice that 1 - e^2 = 1 such that a^2 = b^2.
    The polar equation therefore reduces to r = a with the pole at (ae,0) in Cartesian coordinates. This is clearly a circle centred on the focus.

    Parabola
    The choice of (ae,0) and x = a/e for the focus and directrix respectively means that the equation does not apply to a parabola. This is because e = 1 for a parabola, and the focus would be on the directrix. However, if you just change the focus to (a,0) and the directrix to x = -a then you can use the same method.

    r = e(2a + rcos\theta) but e = 1 so r = 2a + rcos\theta
    We rearrange to find r = \frac{2a}{1 - cos\theta}
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    (Original post by Brister)
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    Fully agree to everything you say, that's a great response, thank you .
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    (Original post by Ateo)
    Fully agree to everything you say, that's a great response, thank you .
    No problem, it was pretty interesting to think about
 
 
 
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