parametric to cartesian equations

Apr 2010
15
0
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!
 
Jun 2009
806
275
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 \(\displaystyle \cos^2(t) + \sin^2(t) = 1\)

.
 
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Dec 2009
872
381
1111
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.

Hope this will help you.
 
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