Easy ellipse question

  • Thread starter winbacker
  • Start date
  • #1
13
0
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?
 

Answers and Replies

  • #2
354
0
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?
 
  • #3
13
0
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...
 
  • #4
354
0
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]​
 

Related Threads on Easy ellipse question

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
932
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
8
Views
723
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
Top