1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$
    – Mittens
    May 25, 2022 at 0:36

1 Answer 1

2
$\begingroup$

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.
$\endgroup$
2
  • $\begingroup$ thank you so much! $\endgroup$
    – DanielG
    May 25, 2022 at 7:10
  • $\begingroup$ I'm glad I could help. $\endgroup$ May 25, 2022 at 7:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .