Parametric equations … ellipse?

Question

The ellipse ((x^2)/(2^2) + (y^2)/(3^2))=1 can be drawn with parametric equations. If x=rcos(t). Then r=? and y=?

Answer

((x^2)/(2^2) + (y^2)/(3^2))=1

How to define our x,y and r?

The “idea” is to compare it and “equalize” appropriately to the tigonometric identity below :

[cos x]^2 + [sin x]^2 = 1

Thus, “equating” we get,
((x^2)/(2^2) = [cos x]^2 and take ‘positive roots’
x/2 = cos x
x = 2 cos x

and,
(y^2)/(3^2)) = [sin x]^2
y/2 = sin x
y = 2 sin x

That is our parametric equations are :
x = 2 cos x
y = 2 sin x

I understand that the “traditional” standard equations for the ellipse are in x,y, and r terms… i just showed how they got it!

r = SQRT ( x^2 + y^2 ) and r>0 by definition.
r = SQRT (4)*( [cos x]^2 + [sin x]^2 )
r = SQRT (4) because ( [cos x]^2 + [sin x]^2 = 1 Trig identity )
r = 2. since r>0.