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Thread: "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 $\displaystyle (x^2 + y^2)^2=8(x^2-y^2)$ where the tangent line is horizontal. I've graphed it, i know the answer is $\displaystyle (-1.75,1),(1.75,1),(-1.75,-1),(1.75,-1)$ 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 $\displaystyle (x^2 + y^2)^2=8(x^2-y^2)$ where the tangent line is horizontal. I've graphed it, i know the answer is $\displaystyle (-1.75,1),(1.75,1),(-1.75,-1),(1.75,-1)$ but how do u get that mathematically?
    Your points are close, but not quite.

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

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

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

    $\displaystyle 2(x^2 + y^2) (2x) = 8 (2x)$

    $\displaystyle 4x^3 + 4xy^2 = 16x$

    $\displaystyle x^2 + y^2 = 4$

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

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

    Thus
    $\displaystyle (x^2 + (4 - x^2))^2=8(x^2-(4 - x^2))$

    You can solve this for the x and y values.

    -Dan
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