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Math Help - parametric to cartesian equations

  1. #1
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    parametric to cartesian equations

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

    <br />
x = 2 + 4\cos t<br />

    <br />
y = 4 + 6\sin t<br />

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

    <br />
\sin t = \frac {y}{24}<br />

    Any help would be appreciated!
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  2. #2
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    Quote Originally Posted by kevin11 View Post
    Find the equivalent Cartesian equation for the parametric equations given below. Which curve does this represent?

    <br />
x = 2 + 4\cos t<br />

    <br />
y = 4 + 6\sin t<br />

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

    <br />
\sin t = \frac {y}{24}<br />

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

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

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

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

    .
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  3. #3
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    Quote Originally Posted by kevin11 View Post
    Find the equivalent Cartesian equation for the parametric equations given below. Which curve does this represent?

    <br />
x = 2 + 4\cos t<br />

    <br />
y = 4 + 6\sin t<br />

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

    <br />
\sin t = \frac {y}{24}<br />

    Any help would be appreciated!
    Dear kevin11,

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

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

    Using the trignometric identity 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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