How do I find the horizontal tangents to the curve $x^3 + y^3 = 6xy$?

Question

How do I find the horizontal tangents to the curve $x^3 + y^3 = 6xy$? James Stewart in section $3.6$ of the 7th edition (on page $167$) shows in a straightforward way that there's a horizontal tangent at the point $(2^{4/3}, 2^{5/3})$ and I can follow the argument. But he restricted himself to the first quadrant, which makes things straightforward. I'm curious as to what is the problem that might happen at the point $(0,0)$. In a certain way, it seems that we could slice the curve into many different functions and there could be a horizontal tangent at $(0,0)$, but perhaps the problem is more complicated than that. (For instance, I'm wondering whether having a crossing at $(0,0)$ is a problem for a derivative there.)

Symbolically, it seems unclear what to deduce because we get a division by zero. The derivative turns out to be $$y' = \frac{2y - x^2}{y^2 - 2x}.$$ So, at $(0,0)$, we get $y' = 0 /0$, which seems to mean that we cannot deduce anything from this. But, at the same time, my instinct says that if there were a derivative there, I would expect this quotient not to be indeterminate, so I'm inclined to say that the derivative does not exist there.

I'm looking for an answer as to whether there is or there isn't a horizontal tangent there. If there isn't, I'm looking for a clear argument as to why. If you know a reference where I can read more about this curve and its problems specifically, I'd appreciate the information.

Answer 1

It is the equation of an algebraic curve passing through the origin. It is known that the set of tangents to an algebraic curve at the origin is given by the homogeneous part of lowest total degree in the equation.

Hence, we can get the set of tangents to the algebraic curve $\;x^3+y^3=6xy\;$ through the equation $\,6xy=0\,.$

But , $\;\;6xy=0\;\iff x=0\;\lor\;y=0\,.$

So there are two tangents at the point $O(0,0)$ and their equations are :

$x=0\;\;$ ($y$-axis) $\;\;$ and $\;\;y=0\;\;$ ($x$-axis) .

Answer 2

According to the plot from Wolfram Alpha, $x^3 + y^3 = 6xy $ is a loop which passes through the origin twice, once with slope zero and once with infinite slope. The curve is symmetric around $y=x$ since, if $(x, y)$ is a solution, so is $(y, x)$.

If $x^3 + y^3 = 6xy $, then $3x^2 dx+3y^2 dy =6(x dy+ y dx) $ or $x^2 dx+y^2 dy =2(x dy+ y dx) $ so $(x^2-2y) dx+(y^2-2x) dy =0 $ or $\dfrac{dy}{dx} =-\dfrac{x^2-2y}{y^2-2x} $.

If $\dfrac{dy}{dx} =0 $ then $x^2=2y$ so $x^3 + (x^2/2)^3 = 6x(x^2/2) $ or $x^3+x^6/8=3x^3 $ or $x^6=16x^3 $ so $x=0$ or $x^3=16 $.

If $x =16^{1/3} =2\cdot 2^{1/3} \approx 2.52 $ then $y =2\cdot 2^{2/3} \approx 3.175 $.

$x^3+y^3 =16+32 =48 $ and $6xy =6(2\cdot 2^{1/3})(2\cdot 2^{2/3}) =6(4\cdot 2) =48 $ so this is on the curve.

If $\dfrac{dx}{dy}=0 $ (vertical tangent) then $y^2=2x$ so $(y^2/2)^3+y^3=6y(y^2/2) $ or $y^6/8+y^3=3y^3 $. This is the same as before with $x$ and $y$ swapped.

Answer 3

The curve is a Folium of Decartes. The derivative is $$\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}$$ It follows that a tangent is horizontal when $2y-x^2=0$ so the required tangents are solutions of the system $$x^3+y^3=6xy\\2y=x^2$$ An obvious one is $\boxed{y=0}$ and the other is $\boxed{y=\dfrac{\sqrt[3]{256}}{2}}$

Answer 4

In polar coordinates (substituting $x = r\cos\theta$ and $y = r\sin\theta$), your equation can be expressed as:

$$r = \frac{6\cos\theta\sin\theta}{\cos^3\theta + \sin^3\theta}$$

Note that there are two ways to reach the origin:

• $\cos \theta = 0$. IOW, $\theta$ is an odd multiple of $\frac{\pi}{2}$.
• $\sin \theta = 0$. IOW, $\theta$ is a multiple of $\pi$.

If you plot the curve on a graph, you see that it intersects itself there.

The slope of the tangent line is:

$$\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x} = \frac{2r\sin\theta - r^2\cos^2\theta}{r^2\sin^2\theta -2r\cos\theta} = \frac{2\sin\theta - r\cos^2\theta}{r\sin^2\theta - 2\cos\theta}$$

When $r = 0$:

$$\frac{dy}{dx} = \frac{2\sin\theta}{- 2\cos\theta} = -\tan\theta$$

The indeterminate form $0/0$ is avoided by having previously cancelled an $r$ from the fraction. Now we just need to plug in $\theta$:

• When it's an odd multiple of $\frac{\pi}{2}$, then $\frac{dy}{dx} = \pm \infty$, so the tangent line is vertical.
• When it's a multiple of $\pi$, then $\frac{dy}{dx} = 0$, so the tangent line is horizontal.

Thus, both $x = 0$ and $y = 0$ are tangent lines.