# Equation for an ellipse

1. Oct 24, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
How would you derive the equation for an ellipse from the parametrization:

x = a cos(t)
y= b sin(t)

If I solve for t and set them equal, I get:

arccos(x/a) = arcsin(x/a)

which looks nothing like the usual formula:

x^2/a^2 + y^2/b^2 = 1

?
2. Relevant equations

3. The attempt at a solution

2. Oct 24, 2007

### robphy

don't focus on t....
try some reverse engineering... how do you know your equations for x and for y describe an ellipse?

3. Oct 24, 2007

### Integral

Staff Emeritus
Think of some trig identies which might look like the typical cartesian fuction for an eclipse. Look at what you have, look at where you need to go. Can you see a path?

4. Oct 24, 2007

### ehrenfest

Clearly if you plug that into x^2/a^2+ y^2/b^2=1 and use s^2 +c^2 = 1 it works, but I just wanted to know how you would get it from arccos(x/a) = arcsin(x/a), pretending, you do not know the traditional equation of an ellipse.

5. Oct 24, 2007

### robphy

arccos(x/a)=arcsin(y/b).
So, in order to isolate one of the variables, one would probably try to write (say) arccos(x/a) in the form: arcsin( f(x) ). In the end, for this problem, you'll certainly return to cos^2(t)+sin^2(t)=1... which you may already know before knowing the traditional non-parametric form of the ellipse.
To see what f(x) should be, you might write the x equation as
x=a*sqrt(1-sin^2 t), then solve for t.

Last edited: Oct 24, 2007
6. Oct 24, 2007

### Dick

sIn(arccos(x))=cos(arcsin(x))=sqrt(1-x^2).

7. Oct 25, 2009

### trailblazer95

Hi,

Does X = a sin(A); Y = b sin(A+B)

give an ellipse equation.

8. Oct 25, 2009

### HallsofIvy

Staff Emeritus
Is A supposed to be the parameter? Is B a constant?

9. Oct 25, 2009

### symbolipoint

Look back at Integral's post #3. Solve for cos(t) and sine(t) in your system of equations. Remember the identity cos2(t)+sin2(t)=1 ?