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    Name:  Parametric Equations Question.jpg
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    Can anyone help me with part e of this question? I've got an idea of where to start but I'm not 100% sure if I'm correct or what to do next.
    The answer is xy + 3y + 5x = -7
    I'm thinking the xy = -7 bit comes from multipling the two equations together, but where does the 3y and 5x come from? Do you have to do some sort of simultaneous equation?

    Thanks,
    Lauren

    P.S. This question is from the C3/C4 Edexcel Revision Workbook. I find the Revision Guide and Workbooks extremely useful and would recommend to anyone doing AS or A2 edexcel maths.
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    Not sure if it's clear in the image. The equations are x = 4e^(-2t) - 3 and y = 2e^(2t) - 5
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    (Original post by Lozzie97)
    Not sure if it's clear in the image. The equations are x = 4e^(-2t) - 3 and y = 2e^(2t) - 5
    Since you want to eliminate the parameter, I'd get the terms involving the parameter by themselves on one side of the equation, and then multiply the two together, eliminating t.
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    (Original post by ghostwalker)
    Since you want to eliminate the parameter, I'd get the terms involving the parameter by themselves on one side of the equation, and then multiply the two together, eliminating t.
    Ah okay. I get that! Thanks! I knew it was something to do with multiplying it out. Fingers crossed I remember this in the actual exam!
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    Hint: Set each equation equal to t then solve.
    Solution in spoiler
    Spoiler:
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    x=2e^{2t}-5 so t=\frac{1}{2}\ln\left(\frac{x+5}  {2}\right)
    y=4e^{-2t}-3 so t=-\frac{1}{2}\ln\left(\frac{y+3}{4  }\right)
    \frac{1}{2}\ln\left(\frac{x+5}{2  }\right)=-\frac{1}{2}\ln\left(\frac{y+3}{4  }\right)
    \ln\left(\frac{x+5}{2}\right)=\l  n\left(\frac{4}{y+3}\right)
    (x+5)(y+3)=8
    So,
    xy+3x+5y=-7


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    (Original post by Cryptokyo)
    Hint: Set each equation equal to t then solve.
    Solution in spoiler
    Spoiler:
    Show
    x=2e^{2t}-5 so t=\frac{1}{2}\ln\left(\frac{x+5}  {2}\right)
    y=4e^{-2t}-3 so t=-\frac{1}{2}\ln\left(\frac{y+3}{4  }\right)
    \frac{1}{2}\ln\left(\frac{x+5}{2  }\right)=-\frac{1}{2}\ln\left(\frac{y+3}{4  }\right)
    \ln\left(\frac{x+5}{2}\right)=\l  n\left(\frac{4}{y+3}\right)
    (x+5)(y+3)=8
    So,
    xy+3x+5y=-7


    It's easier to consider x+5 and y+3 in this case.
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    Ah yes didn't spot that :P !!!
    (Original post by B_9710)
    It's easier to consider x+5 and y+3 in this case.
 
 
 
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