Cusps versus vertical tangents for the graphs $x^{2/3}+y^{2/3}=4$ and $x^{4/3}+y^{4/3}=16$

Question

I have managed to find the domain, range and derivative at the endpoints of the domain and range for both graphs. I have confirmed my sketches are accurate with a graphing calculator.

My questions are theoretical:

• for the first graph $x^{2/3}+y^{2/3}=4$:

1. are there cusps at $(0,8)$ and $(0, -8)$ and
2. are there horizontal tangents at $(8,0)$ and $(-8,0)$?

• for the second graph $x^{4/3}+y^{4/3}=16$:

3. are there horizontal tangents at $(0,8)$ and $(0,-8)$ and
4. are there cusps or vertical tangents at $(8,0)$ and $(-8,0)$?

I’m only up to high school math. Thank you for keeping it simple.

Answer 1

A) Given that an equation for $C_1: x^{2/3}+y^{2/3}=4$ is $(x^2+y^2-8^2)^3+27x^2y^2\cdot 8^2=0:$

1. Following the elements of least degree at $(0,\pm 8):$ $\begin{align}(x^2+(y\mp2^3)^2-2^{3})^3+\color{red}{3^3x^2}(y\mp \color{red}{2^3})^\color{red}{2}\cdot \color{red}{2^{6}}&=0\\(x^2+(\ldots \mp 2y\cdot (2^3)+\color{blue}{(2^3)^2})-\color{blue}{2^{6}})^3+3^3x^2(y\mp 2^3)^2\cdot 2^{6}&=0\\\ldots \mp 3^3 \cdot 2^{6+6}x^2&=0\end{align}$

So substituting back we get the tangent cone $\mp 3^3\cdot 2^{12}x^2=0$

2. Following the elements of least degree at $(\pm 8,0):$ $\begin{align}(y^2+(x\mp2^3)^2-2^{3})^3+\color{red}{3^3y^2}(x\mp \color{red}{2^3})^\color{red}{2}\cdot \color{red}{2^{6}}&=0\\(y^2+(\ldots \mp 2x\cdot (2^3)+\color{blue}{(2^3)^2})-\color{blue}{2^{6}})^3+3^3y^2(x\mp 2^3)^2\cdot 2^{6}&=0\\\ldots \mp 3^3 \cdot 2^{6+6}y^2&=0\end{align}$

So substituting back we get the tangent cone $\mp 3^3\cdot 2^{12}y^2=0$

In conclusion $(0,\pm 8), (\pm 8,0)$ are ordinary cusps for the astroid $C_1$.

B) Given that an equation for $C_2: x^{4/3}+y^{4/3}=16$ is $(x^4+y^4-8^4)^3+27x^4y^4\cdot 8^4=0:$

1. Following the elements of least degree at $(0,\pm 8):$ $\begin{align}(x^4+(y\mp2^3)^4-2^{12})^3+3^3x^4(y\mp 2^3)^4\cdot 2^{12}&=0\\(x^4+(\ldots \color{red}{\mp 4y\cdot (2^3)^3}+\color{blue}{(2^3)^4})-\color{blue}{2^{12}})^3+3^3\color{blue}{x^4}(y\mp 2^3)^4\cdot 2^{12}&=0\\\ldots \mp(y\cdot 2^{11})^3&=0\end{align}$

So substituting back we get the tangent cone $\mp 2^{33}(y\mp 8)^3=0$

2. Following the elements of least degree at $(\pm 8,0):$ $\begin{align}(y^4+(x\mp2^3)^4-2^{12})^3+3^3y^4(x\mp 2^3)^4\cdot 2^{12}&=0\\(y^4+(\ldots \color{red}{\mp 4x\cdot (2^3)^3}+\color{blue}{(2^3)^4})-\color{blue}{2^{12}})^3+3^3\color{blue}{y^4}(x\mp 2^3)^4\cdot 2^{12}&=0\\\ldots \mp(x\cdot 2^{11})^3&=0\end{align}$

So substituting back we get the tangent cone $\mp 2^{33}(x\mp 8)^3=0$

In conclusion $(0,\pm 8), (\pm 8,0)$ have tangents (horizontal for $(0,\pm 8)$, vertical for $(\pm 8,0)$) with higher contact for $C_2.$