Problem:
Find the coordinates of any point on the curve of $y^2-4xy=x^2+5$ for which the tangent is horizontal.
When I solve $\frac{dy}{dx}=0$, I end up getting $x=-2y$. So then I was thinking that I can put in any number for $x$ and get a corresponding $y$ where $(x,y)$ is a point where the slope is zero. So I plugged in $1$ for $x$ and got $-\frac{1}{2}$ for $y$, which is the point $(1,-\frac{1}{2})$. But this point can't be a point where the slope is zero because graphically the slope is zero only at $(2,-1)$ and $(-2,1)$. What am I missing?
Best Answer
After getting $x =-2y$ you also have to take into account the fact that $y^{2}-4xy=x^{2}+5$ because you have to find points ** on the curve** at which the derivative is $0$. If you solve these two equations simultaneously you will get the answer.