Deriving the Equation for an Ellipse from Parametrization

[ehrenfest]

Homework Statement

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

?

Homework Equations

The Attempt at a Solution

 

[robphy]

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

 

[Integral]

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?

 

[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.

 

[robphy]

Of course, with your approach, your starting point is
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.

 

[Dick]

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

 

[trailblazer95]

Hi,

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

give an ellipse equation.

 

[HallsofIvy]

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

 

[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 ?