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Math Help - help with proof

  1. #1
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    help with proof

    for the curve C x3 + xy + 2y3 =k , the y co-ordinates of the tangent parallel to the y axis is given by

    216y6 +4y3 +k = 0


    Show that k<= 1/54
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  2. #2
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    Re: help with proof

    Let m = y^3, so the second equation becomes


    216m^2 + 4m + k = 0


    The discriminant of this quadratic is 16 - 4(216)(k) = 16 - 864k


    For the quadratic to even have real solutions for m, the discriminant must be non-negative, i.e. 16 - 864k \ge 0 \Rightarrow k \le \frac{1}{54}. No calculus needed.
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  3. #3
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    Re: help with proof

    but its a tangent shouldnt you would get a repeated root because its a tangent so should you use the the discrm = 0 to find k than puttin >=0
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  4. #4
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    Re: help with proof

    Can you prove that? Note that the quadratic is now in terms of m, not y.
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  5. #5
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    Re: help with proof

    thanks so you cannot assum repeated for m, thanks
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  6. #6
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    Re: help with proof

    one more find the possible values of k for which the line x=-6 is a tangent to C
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  7. #7
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    Re: help with proof

    No problem
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  8. #8
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    Re: help with proof

    For the other question, differentiate both sides of the original equation:


    3x^2 + (y + x\frac{dy}{dx}) + 6y^2 \frac{dy}{dx} = 0


    \frac{dy}{dx} (x + 6y^2) + 3x^2 + y = 0


    \frac{dy}{dx} = -\frac{3x^2 + y}{x + 6y^2}


    If the line x = -6 is a tangent at the point (x,y), it means that x = -6 and dy/dx is undefined, i.e. the denominator of dy/dx is zero. This occurs only when y = \pm 1. So try substituting (1,6) and (-1,6) into your original equation and find k.
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