Solve for y within a parametric equation

Question

The ellipse

$$\frac{x^2}{2^2} + \frac{y^2}{3^2} = 1$$

can be drawn with parametric equations. Assume the curve is traced clockwise as the parameter increases.

If $x=2\cos t $

then $y =$ _____

I got $y=(9-(9x^2)/4)^{1/2}$ and then substitute $x$ for $2\cos t$ and got $y=(9-(9(2\cos t)^2)/4)^{1/2}$. What am I doing wrong?

Answer 1

you are doing nothing wrong. Just try to simplify your "final" expression you would find it is $3\sin(t)$ or $-3\sin(t)$.

Answer 2

Hint $:$ Plug the value $x=2 \cos t$ in the equation $\frac {x^2} {2^2} + \frac {y^2} {3^2} =1.$

You will end up with

$$\cos^2 t + \frac {y^2} {3^2} = 1 \implies \frac {y^2} {3^2} = 1 - \cos^2 t \implies y^2 = 3^2 \sin^2 t \implies y = \pm 3 \sin t.$$