# Find the points on the curve where the tangent is horizontal

Question. Given $$y^{2}=x^{3}+ax+b$$, find the points on the curve where the tangent line is horizontal.

Attempt. Let $$f(x,y)=x^{3}-y^{2}+ax+b=0$$

The tangent is horizontal at points where the gradient is vertical.

The gradient is vertical if and only if $$\begin{cases} 3x^{2}+a=0\\ y^{2}=x^{3}+ax+b \end{cases}$$

I got entangled with the parameters, and I'm not sure how to continue forward.

• Use implicit function theorem $\frac{dy}{dx}=-\frac{\partial_yf}{\partial_xf}=0$ for the horizontal tangent, and $\frac{dx}{dy}=-\frac{\partial_xF}{\partial_yf}=0$ for the vertical line May 25, 2022 at 0:36

Actually, the gradient is vertical if and only if$$\left\{\begin{array}{l}3x^2+a=0\\-2y\ne0(\iff y\ne0)\\y^2=x^3+ax+b.\end{array}\right.$$
• if $$a>0$$, the system has no solutions, since the first equation has no solutions;
• if $$a=0$$ and $$b<0$$, the system has no solutions, since the only solution of the first equation is $$x=0$$;
• if $$a=b=0$$, the system has no solutions, since the only solution of the first and third equation is $$x=y=0$$;
• if $$a=0$$ and $$b>0$$, then the solutions are $$\left(0,\pm\sqrt b\right)$$;
• if $$a<0$$, then the first equation has two roots: $$\pm\sqrt{-\frac a3}$$ and then$$x^3+ax+b=x(x^2+a)+b=\pm\sqrt{-\frac a3}\times\frac{2a}3+b.$$Now, you shall have to consider the values that the numbers $$\pm\sqrt{-\frac a3}\times\frac{2a}3+b$$ can take. For instance, if both of them are greater than $$0$$, then there are four points at which the tangent is horizontal. If$$\sqrt{-\frac a3}\times\frac{2a}3+b<0<-\sqrt{-\frac a3}\times\frac{2a}3+b,$$then there are two such points. And so on.