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Thread: polar form of ellipse

  1. #1
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    polar form of ellipse

    The problem is to prove that the following curve is an ellipse

    r=\frac{2a}{3+2\cos\theta}

    I've looked about and have found the equation of the ellipse in polar form, but I can't find it's derivation or proof anywhere.

    Anyway, I started this problem by subbing in

    r = \sqrt{x^2+y^2}

    and

    x = r\cos\theta

    But I ended up with a horrible mess and couldn't spot how to get it into the form \frac{x^2}{a^2}+\frac{y^2}{b^2}=1

    Any help or guidance would be much appreciated

    Stonehambey
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  2. #2
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    Hello, Stonehambey!

    Prove that the following curve is an ellipse:

    . . r\:=\:\frac{2k}{3+2\cos\theta}

    You have the right idea, but I've learned that it's better to simplify first.


    We have: . r(3 + 2\cos\theta) \:=\:2k \quad\Rightarrow\quad 3r + 2r\cos\theta \:=\:2k \quad\Rightarrow\quad 3r \:=\:2k - 2r\cos\theta

    Substitute now: . 3\underbrace{r} \;=\;2k - 2\underbrace{r\cos\theta}
    . . . . . . . . . . . . . . \downarrow\qquad\qquad\qquad\;\; \downarrow
    . . . . . . . . . . 3\sqrt{x^2+y^2} \;\;=\;\; 2k - 2x


    Square both sides: . 9(x^2+y^2) \;=\;(2k-2x)^2

    . . . . . . 9x^2 + 9y^2 \;=\;4k^2 - 8kx + 4x^2 \quad\Rightarrow\quad 5x^2 + 8kx + 9y^2 \;=\;4k^2


    Complete the square: . 5\left(x^2 + \tfrac{8}{5}x\right) + 9y^2 \;=\;4k^2

    . . 5\left(x^2 + \tfrac{8}{5}kx + {\color{red}\tfrac{16}{25}k^2}\right) + 9y^2 \;=\;4k^2 + {\color{red}\tfrac{16}{5}k^2} \quad\Rightarrow\quad 5\left(x + \tfrac{4}{5}k\right)^2 + 9y^2 \;=\;\tfrac{36}{5}k^2


    Divide by \tfrac{36}{5}k^2\!:\quad\frac{5(x+\frac{4}{5}k)^2}  {\frac{36}{5}k^2} + \frac{9y^2}{\frac{36}{5}k^2} \;=\;1

    And we have: . \frac{(x+\frac{4}{5}k)^2}{\frac{36}{25}k^2} + \frac{y^2}{\frac{4}{5}k^2} \;=\;1


    This an ellipse with: . center \left(\text{-}\tfrac{4}{5}k,\:0\right),\quad a \:=\: \tfrac{6}{5}k,\quad b \:=\:\tfrac{2}{\sqrt{5}}k

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  3. #3
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    Quote Originally Posted by Stonehambey View Post
    The problem is to prove that the following curve is an ellipse

    r=\frac{2a}{3+2\cos\theta}

    I've looked about and have found the equation of the ellipse in polar form, but I can't find it's derivation or proof anywhere.

    Anyway, I started this problem by subbing in

    r = \sqrt{x^2+y^2}

    and

    x = r\cos\theta

    But I ended up with a horrible mess and couldn't spot how to get it into the form \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
    There are two ways of doing this: geometric or analytic.

    Geometrically, an ellipse is defined as the locus of a point which moves so that its distance from a fixed line (the directrix) is a constant multiple (the eccentricity, which must be less than 1) of its distance from a fixed point (the focus). In this case, you can write the equation as r = \tfrac23(a-x). This says that r (the distance of the point [r,θ] from the origin) is 2/3rds of its distance from the line x=a. So the locus is an ellipse, with focus at the origin, directrix the line x=a, and eccentricity 2/3.

    To solve the problem analytically, start from the equation in the form 3r = 2(a-x). Square both sides: 9(x^2+y^2) = 4(a-x)^2 = 4a^2-8ax+4x^2. Rearrange this as 5x^5+8ax+9y^2 = 4a^2. Complete the square: 5(x+\tfrac45a)^2 + 9y^2 = \tfrac{36}{25}a^2  = (\tfrac65a)^2. Finally, divide both sides by (\tfrac65a)^2 and you get the equation in the form \frac{(x+\frac45a)^2}{(\frac65a)^2} + \frac{y^2}{(\frac2{\sqrt5}a)^2} = 1. Again, this is the equation of an ellipse. It is not centred at the origin, but at the point (-\tfrac45a,0). Its semi-axes are \tfrac65a and \tfrac2{\sqrt5}a.

    Edit. Soroban got there first with the analytic version!
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