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Math Help - "hard" implict tangent question

  1. #1
    Senior Member polymerase's Avatar
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    "hard" implict tangent question

    Find all the ponts on the curve (x^2 + y^2)^2=8(x^2-y^2) where the tangent line is horizontal. I've graphed it, i know the answer is (-1.75,1),(1.75,1),(-1.75,-1),(1.75,-1) but how do u get that mathematically?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    Find all the ponts on the curve (x^2 + y^2)^2=8(x^2-y^2) where the tangent line is horizontal. I've graphed it, i know the answer is (-1.75,1),(1.75,1),(-1.75,-1),(1.75,-1) but how do u get that mathematically?
    Your points are close, but not quite.

    Take the derivative:
    2(x^2 + y^2) \left ( 2x + 2y\frac{dy}{dx} \right ) = 8 \left ( 2x - 2y \frac{dy}{dx} \right )

    The slope of the tangent is horizontal when \frac{dy}{dx} = 0

    Sooooo....
    2(x^2 + y^2) ( 2x + 2y\cdot 0 ) = 8 ( 2x - 2y \cdot 0 )

    2(x^2 + y^2) (2x) = 8 (2x)

    4x^3 + 4xy^2 = 16x

    x^2 + y^2 = 4

    So the points where we have a 0 slope tangent are the points where the original function intersects the circle x^2 + y^2 = 4.

    So y^2 = 4 - x^2 is our condition we insert into the original relation.

    Thus
    (x^2 + (4 - x^2))^2=8(x^2-(4 - x^2))

    You can solve this for the x and y values.

    -Dan
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