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With the equation $x^2+2xy-3y^2+16=0$, I need to find the coordinates of the points on the curve where $\frac{dy}{dx} = 0$. I think I have correctly used implicit differentiation to get $\frac{dy}{dx} = \frac{-x-y}{x-3y}$. And the multiplied by the denominator and subsituted zero to get, $0 = -x-y$, however, I'm not sure where to go from here.

If anyone could help that would be great. Thanks.

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At the point where $(x,y)$ where $dy/dx=0$, it follows that $y=-x$. To find $x$ and $y$, note that $$ x^2+2x(-x)-3(-x)^2+16=0 $$ Solve for $x$ and substitute into $y=-x$ to find $y$.

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Thanks for all the help. Doing what you've said I have the coordinates as (2,-2) and (-2,2).

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Hint: The point is on the curve, so it also satisfies $x^2+2xy-3y^2+16=0$.

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just use implicit differentiation for the equation you stated for the curve $y=y(x)$ and you get $2x +2x*y'(x)+2y(x)-6y(x)y'(x)=0$. Now solving after $y'=\frac{dy}{dx}$ you get
$y'(x)*(2x-6y(x))+2x+2y(x)=0$ and hence $y(x)=-x$

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