# parametric to cartesian equations

#### kevin11

Find the equivalent Cartesian equation for the parametric equations given below. Which curve does this represent?

$$\displaystyle x = 2 + 4\cos t$$

$$\displaystyle y = 4 + 6\sin t$$

So far all I've done is isolate x and y by itself (not sure if this is the write course of action).
$$\displaystyle \cos t = \frac {x}{8}$$

$$\displaystyle \sin t = \frac {y}{24}$$

Any help would be appreciated!

#### sa-ri-ga-ma

Find the equivalent Cartesian equation for the parametric equations given below. Which curve does this represent?

$$\displaystyle x = 2 + 4\cos t$$

$$\displaystyle y = 4 + 6\sin t$$

So far all I've done is isolate x and y by itself (not sure if this is the write course of action).
$$\displaystyle \cos t = \frac {x}{8}$$

$$\displaystyle \sin t = \frac {y}{24}$$

Any help would be appreciated!
No. Your simplification is not correct. It should be

$$\displaystyle \cos(t) = \frac{x-2}{4}$$

$$\displaystyle \sin(t) = \frac{y-4}{6}$$

Now use identity $$\displaystyle \cos^2(t) + \sin^2(t) = 1$$

.

kevin11

#### Sudharaka

Find the equivalent Cartesian equation for the parametric equations given below. Which curve does this represent?

$$\displaystyle x = 2 + 4\cos t$$

$$\displaystyle y = 4 + 6\sin t$$

So far all I've done is isolate x and y by itself (not sure if this is the write course of action).
$$\displaystyle \cos t = \frac {x}{8}$$

$$\displaystyle \sin t = \frac {y}{24}$$

Any help would be appreciated!
Dear kevin11,

Your method of isolation of sine and cosine terms was incorrect.

You should get, $$\displaystyle sint=\frac{y-4}{6}$$ and $$\displaystyle cost=\frac{x-2}{4}$$

Using the trignometric identity $$\displaystyle sin^{2}t+cos^{2}t=1$$ you can eliminate t to obtain the cartesian equation.