There is this practice problem that asks to determine the points at which the graph of $y^4=y^2-x^2$ has a horizontal tangent.

So I did implicit differentiation to find that

$$\displaystyle\frac{dy}{dx} = \frac{-x}{2y^3-y}$$

To find the horizontal tangent, I set $\frac{dy}{dx}=0$ and solved for $x$:

$$\begin{align} \displaystyle\frac{dy}{dx} &= 0 \\ \frac{-x}{2y^3-y} &= 0 \\ -x &= 0 \\ x &= 0 \end{align}$$

Then I substituted $x=0$ into the equation of the curve:

$$\begin{align} y^4&=y^2-(0)^2 \\ 0 &= y^4 - y^2\\ 0 &= y^2(y+1)(y-1) \\ y&=-1,\,0,\,1 \end{align}$$

I concluded that the points $(0,0)$, $(0,-1)$, and $(0,1)$ were the points with a horizontal tangent.

However, when I graphed this using Desmos, it turns out that the point at $(0,0)$ did not look like it has horizontal tangent.

Graph of y^4=y^2-x^2

How would I have been able to figure this out without graphing it?

  • 2
    $\begingroup$ At $y = 0$ derivative is not defined! Simply look at your expression for derivative for that. $\endgroup$ – SJ. Nov 19 '17 at 4:25
  • $\begingroup$ Ah! did not even consider that. $\endgroup$ – user502227 Nov 19 '17 at 5:26

One of the answers is $(0,0)$. However, at this point, $\frac{dy}{dx}$ attains $\frac{0}{0}$ form. So, I think you should not consider the point $(0,0)$.


If $y^4=y^2-x^2 $, then diffing implicitly, $4y^3y' =2yy'-2x $ or $2y^3y' =yy'-x $.

If $y' = 0$, then $x = 0$.

Putting this in, $y^4 = y^2$ so the possible values are $y = 0, \pm 1$.

At $x=y=0$, suppose $y' = c$. For small $x$ and $y$, $y \approx cx$ so $c^4x^4 \approx c^2x^2-x^2$ or, dividing by $x^2$, $c^4 x^2 \approx c^2-1$. Since $c^4x^2$ is small, $c^2-1$ must also be small so $c^2 \approx 1$ so $c \approx \pm 1$.

This means that, at $(0, 0)$, $y' = \pm 1$.


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