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Math Help - Elimination technique questions that frankly, I don't understand.

  1. #1
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    Elimination technique questions that frankly, I don't understand.

    1) Eliminate t to give an equation that relates x and y.
    x=(e^(2t))+9 and y=e^(7t)

    2) Eliminate t to give an equation that relates x and y.
    x=tan(t) and y=sec^2(t)-3

    3) Show that this equation represents a circle by rearranging it into the centre-radius form of the equation of a circle. State the coordinates of the center and the radius of the circle:

    21y^2+19x^2+83=4x-2x^2-84y



    LOLWUT?! Help would be soo much appreciated.
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  2. #2
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    Quote Originally Posted by katrina782 View Post
    1) Eliminate t to give an equation that relates x and y.
    x=(e^(2t))+9 and y=e^(7t)
    1. Solve the 2nd equation for t:

    y=e^{7t}~\implies~7t=\ln(y)~\implies~\boxed{t=\fra  c17 \ln(y)}
    and plug in this term into the first equation:

    x=\left(e^{2 \cdot \frac17 \ln(y)}\right)+9~\implies~x=\sqrt[7]{y^2}+9~\implies~y^2=(x-9)^7\ ,\ x>9

    2) Eliminate t to give an equation that relates x and y.
    x=tan(t) and y=sec^2(t)-3
    This one is for you!

    3) Show that this equation represents a circle by rearranging it into the centre-radius form of the equation of a circle. State the coordinates of the center and the radius of the circle:

    21y^2+19x^2+83=4x-2x^2-84y

    ...
    1. Re-arrange the equation:

    21x^2 -4x+21y^2+84y = -83

    2. Complete the squares:

    21\left(x^2 - \frac4{21} x + \frac{4}{441}\right) +21\left(y^2+4y+4\right) = -83+21 \cdot \frac4{441} + 21 \cdot 4 = \frac{25}{21}

    3. Divide the equation by the leading factor of the brackets on the LHS:

    \left(x-\frac2{21}\right)^2+(y+2)^2=\left(\frac5{21}\right  )^2

    4. Determine the coordinates of the centre and the length of the radius.
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  3. #3
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    You're dope. Thanks
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  4. #4
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    So "you're dope" is different from "you're a dope"?

    I would have done (1) slightly differently, solving only for e^t rather than t itself. From the first equation, [/tex]x= e^{2t}+ 9[/tex], x- 9= e^{2t}= (e^t)^2, e^t= (x-9)^{1/2}. From the second equation, y= e^{7t}= (e^t)^7, e^t= y^{1/7}. Putting those together, (x-9)^{1/2}= y^{1/7}, which, taking both sides to the 14th power, is the same as (x-9)^7= y^2
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