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Easy ellipse question

  1. Mar 25, 2009 #1
    I have the solution to a problem and I need help understanding how the solution was obtained.

    I have to take an equation of an ellipse and transfer it to parametric form.

    The ellipse has the equation: (x+1)^2 + 4y^2 = 4.

    The solution has the ellipse parameterized as follows:

    x+1 = 2cost and y = sint

    Can someone just explain this translation, step by step?
     
  2. jcsd
  3. Mar 25, 2009 #2
    The parametric form of an arbitrary ellipse is

    [tex]\begin{align*}
    x &= \alpha\cos t + c_x\\
    y &= \beta\sin t + c_y
    \end{align*}
    [/tex]

    where [tex](c_x, c_y)[/tex] is its center and [tex]\alpha,\beta[/tex] are constants.

    How would you use this to convert your equation to parametric form? Can you find values for any of [tex]c_x,c_y,\alpha[/tex], or [tex]\beta[/tex] right off the bat?
     
  4. Mar 25, 2009 #3
    well, I assume the center of the ellipse is at (-1,0). Otherwise I am stumped....how do they get 2cost?? And how is y = sint? I am very confused...
     
  5. Mar 26, 2009 #4
    You know what, I just thought of a better way to do this.

    Recall that in parametric form the equation [tex]\frac{x^2}{r^2}+\frac{y^2}{r^2}=1[/tex] is
    [tex]\begin{equation}\tag{1}\begin{align*}
    x &=r\cos t\\
    y &=r\sin t.
    \end{align*}\end{equation}
    [/tex]​

    So, to put your equation
    [tex](x+1)^2 + 4y^2 = 4[/tex]​
    into parametric form, we need to first do some sort of substitution of variables to get it to look like the equation of a circle. In other words, we want to find functions [tex]f[/tex] and [tex]g[/tex] such that if
    [tex]\begin{aligned}
    x &= f(u,v)\\
    y &= g(u,v)
    \end{aligned}[/tex]​
    (for new variables [tex]u[/tex] and [tex]v[/tex]), then
    [tex]\begin{equation}\tag{2}\frac{u^2}{r^2} + \frac{v^2}{r^2} = 1.\end{equation}[/tex]​
    Once we do so, we can put (2) in parametric form as
    [tex]\begin{equation}\tag{3}\begin{aligned}
    u &= r\cos t\\
    v &= r\sin t
    \end{aligned}\end{equation}[/tex]​
    and then (since u will turn out to only depend on x, and v only on y) we can plug in the inverse substitution of (2) to get the parametrization we want.

    Let's start by dividing the equation by 4:
    [tex]\frac{(x+1)^2}{4}+y^2=1.[/tex] ​
    The obvious choice for y is to let [tex]y=v\ (=g(u,v))[/tex].

    x, though, is not quite as simple. However, a major benefit of our choice for y is that since [tex]\frac{(x+1)^2}{4}[/tex] doesn't depend on y, it also doesn't depend on v. Thus, instead of finding a function f(u,v), we just need to find a function f(u)! Additionally, notice that by letting y=u, we have effectively chosen r to be 1. So, to we need to find a function f such that
    [tex]\frac{(f(u)+1)^2}{4}=\frac{u^2}{1}=u^2.[/tex]​
    Solving, we get
    [tex]f(u)=2u-1\ (=x).[/tex]​

    So, to consolidate, we have
    [tex]\begin{equation}\tag{4}\begin{aligned}
    x &= 2u-1\\
    y &= v.
    \end{aligned}\end{equation}[/tex]​
    Plugging the inverse substitution
    [tex]\begin{equation}\begin{aligned}
    u &= \frac{x+1}{2}\\
    v &= y
    \end{aligned}\end{equation}[/tex]​
    (and r=1) into (3), we have
    [tex]\begin{align*}
    \frac{x+1}{2} &= \cos t\\
    y &= \sin t
    \end{align*}[/tex]​
     
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