## 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.

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