Answer
The slope of the curve at $(-2,1)$ is $-1$ and the slope of the curve at $(-2,-1)$ is $1$.
Work Step by Step
$$y^2+x^2=y^4-2x$$
1) Find $dy/dx$:
Differentiate both sides of the equation with respect to $x$: $$2y\frac{dy}{dx}+2x=4y^3\frac{dy}{dx}-2$$
Isolate the terms with $dy/dx$ into one side and solve for $dy/dx$: $$4y^3\frac{dy}{dx}-2y\frac{dy}{dx}=2x+2$$ $$2y^3\frac{dy}{dx}-y\frac{dy}{dx}=x+1$$ $$(2y^3-y)\frac{dy}{dx}=x+1$$ $$\frac{dy}{dx}=\frac{x+1}{2y^3-y}$$
2) The slope of the curve at a point $(a,b)$ is the value of the derivative $dy/dx$ at that point.
- At $(-2,1)$: $$\frac{dy}{dx}=\frac{-2+1}{2\times1^3-1}=\frac{-1}{1}=-1$$
The slope of the curve at $(-2,1)$ is $-1$.
- At $(-2-,1)$: $$\frac{dy}{dx}=\frac{-2+1}{2\times(-1)^3-(-1)}=\frac{-1}{-2+1}=\frac{-1}{-1}=1$$
The slope of the curve at $(-2,-1)$ is $1$.
