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Math Help - parametric trig curve

  1. #1
    Senior Member furor celtica's Avatar
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    parametric trig curve

    find the equation of the normal to the curve x= 2cosALPHA, y = 3sinALPHA at the point where ALPHA= pi/4
    find the coordinates of the point where this normal cuts the curve again.

    i didnt have any problems with the first question; i found the equation of the normal is y = 2x/3 - 2sqrt2/3 + 3sqrt2/2
    but from here its hard to find the point where the curve and the normal intersect again. i tried using 3sinALPHA=2x/3 - 2sqrt2/3 + 3sqrt2/2 and replacing ALPHA with arccos(x/2) but that doesnt make sense does it?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Another approach is to note

    1) (x/2)^2 +(y/3)^2 = 1 using x/2 = cos(a) y/3 = sin(a)


    and y = 2/3 x + 5sqrt(2)/6

    substitute this into 1)

    Still it won't be pretty and I hope you have a computer or calculator to

    solve.

    See attachment where I have used Mathcad to solve the problem
    Attached Thumbnails Attached Thumbnails parametric trig curve-normalline.jpg  
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  3. #3
    Senior Member furor celtica's Avatar
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    hey but the equation ends up being quadratic, which value is the right one? your explanation is a bit confusing. can you show me step by step?
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  4. #4
    MHF Contributor

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    The problem was to find where the curve and the normal line intersect again! Obviously the original point is one solution to that quadratic equation. The other solution is where they intersect again.
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