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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- Oct 17th 2007, 02:57 PMpolymerase"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?

- Oct 17th 2007, 03:28 PMtopsquark
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