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    Q3a). Find the equation of the normal to the curve with parametric equations x=e^t, y=e^t + e^-t, at point P, where t=2
    I managed to get the answer as y=2 but the answer is apparently x=1?

    Thanks in advance
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    (Original post by doglover123)
    Q3a). Find the equation of the normal to the curve with parametric equations x=e^t, y=e^t + e^-t, at point P, where t=2
    I managed to get the answer as y=2 but the answer is apparently x=1?

    Thanks in advance
    Could you provide some working please so we can identify where you went wrong?

    Edit 1: Neither x=1 nor y=2 are going to be right as the normal will be oblique and those lines do not pass through P.

    Edit 2: Did you mean t = 0 and did you find the tangent rather than the normal?
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    (Original post by Mr M)
    Could you provide some working please so we can identify where you went wrong?

    Edit 1: Neither x=1 nor y=2 are going to be right as the normal will be oblique and those lines do not pass through P.

    Edit 2: Did you mean t = 0 and did you find the tangent rather than the normal?
    I'm sorry, I was looking at the q before, t does equal 0.

    dy/dt = e^t-e^-t/e^t
    substitute t=0 dy/dx = 0 therefore gradient of normal = 0 and (1,20
    y-y1=m(x-x1)
    y-2=0
    y=2
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    (Original post by doglover123)
    I'm sorry, I was looking at the q before, t does equal 0.

    dy/dt = e^t-e^-t/e^t
    substitute t=0 dy/dx = 0 therefore gradient of normal = 0 and (1,20
    y-y1=m(x-x1)
    y-2=0
    y=2
    Right firstly you don't mean dy/dt you mean dy/dx. But dy/dx is the gradient of the tangent not the normal. If the tangent is horizontal (this is what you have just found) then the normal will be vertical.
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    (Original post by Mr M)
    Right firstly you don't mean dy/dt you mean dy/dx. But dy/dx is the gradient of the tangent not the normal. If the tangent is horizontal (this is what you have just found) then the normal will be vertical.
    Ahhh I didn't know that! Thanks for clearing things up, also I apologise for the typo's, I should have proof read. Thanks again!
 
 
 
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