# parametric to cartesian equations

• May 23rd 2010, 04:35 PM
kevin11
parametric to cartesian equations
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!
• May 23rd 2010, 04:47 PM
sa-ri-ga-ma
Quote:

Originally Posted by 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!

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

.
• May 23rd 2010, 04:48 PM
Sudharaka
Quote:

Originally Posted by 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!

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.