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Math Help - Vertical Tangent to a Curve

  1. #1
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    Vertical Tangent to a Curve

    I'm having trouble with this problem.

    The tangent to the curve (y^2)-xy+9=0 is vertical when
    (a) y= 0
    (b) y= +/- radical 3
    (c) y= 1/2
    (d) y= +/- 3
    (e) none of these

    I took the derivative of the equation and solved for x' and I got
    x'= (x-2y)/(-y)
    and then I set it to 0 but I get
    x-2y=0
    and I don't know what to do from there.
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  2. #2
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    x - 2y = 0

    x = 2y

    substitute 2y for x in the original equation ... solve for y, and then determine x.
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  3. #3
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    Hello, summermagic!

    The tangent to the curve: y^2-xy+9\:=\:0 is vertical when:

    . . (a)\;y= 0 \qquad (b)\;y= \pm\sqrt{3} \qquad (c)\;y= \tfrac{1}{2} \qquad (d)\;y= \pm3 \qquad (e)\text{ none of these}

    Differentiate implicitly: . 2yy' - xy' - y \:=\:0\quad\Rightarrow\quad y' \:=\:\frac{y}{2y-x}

    The tangent is vertical when the denominator is zero:
    . . 2y - x \:=\:0 \quad\Rightarrow\quad x \:=\:2y

    Substitute: . y^2 - (2y)y + 9 \:=\:0 \quad\Rightarrow\quad -y^2 + 9 \:=\:0

    Therefore: . y^2 \:=\:9 \quad\Rightarrow\quad y \:=\:\pm3 . . . answer (d)

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  4. #4
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    Okay, thank you!
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