Indeed, $p=\frac{2}{3}$ is not special; there is a general construction, for which I define
$$F(x,y,a):= \frac{|x|^q}{a^r} + \frac{|y|^q}{(1-a)^r} - 1.\tag{1}$$
The envelope is found by eliminating $a$ from the system
$$F(x,y,a) = 0, \tag{2}$$
$$\frac{\partial F}{\partial a}(x,y,a) = 0. \tag{3}$$
From $(3)$ we obtain $-r |x|^q a^{-r-1} + r |y|^q (1-a)^{-r-1} = 0$ which, assuming $r \neq 0$, we rearrange to separate $x$ and $y$:
$$\frac{|x|^q}{a^{r+1}} = \frac{|y|^q}{(1-a)^{r+1}}.$$
Taking the $(r+1)$-th root of both sides, and letting $k = \dfrac{q}{r+1}$, gives $\dfrac{|x|^k}{a} = \dfrac{|y|^k}{1-a}$, or
$$|x|^k = a(|x|^k + |y|^k).$$
This gives us explicit expressions for $a$ and $1-a$:
$$a = \frac{|x|^k}{|x|^k + |y|^k},\qquad 1-a = 1 - \frac{|x|^k}{|x|^k + |y|^k} = \frac{|y|^k}{|x|^k + |y|^k}.\tag{4}$$
Substitute $(4)$ back into the original family equation $(1,2)$ to get
$$|x|^q \left( \frac{|x|^k + |y|^k}{|x|^k} \right)^{\!\!r} + |y|^q \left( \frac{|x|^k + |y|^k}{|y|^k} \right)^{\!\!r} = 1,$$
or $$(|x|^k + |y|^k)^r \left( \dfrac{|x|^q}{|x|^{kr}} + \dfrac{|y|^q}{|y|^{kr}} \right) = 1.$$ Recalling that $k = \dfrac{q}{r+1}$, we have $q = kr + k$, which implies $q - kr = k$. Thus
$(|x|^k + |y|^k)^r ( |x|^k + |y|^k) = 1$, or
$$(|x|^k + |y|^k)^{r+1} = 1.$$
Taking the $(r+1)$-th root gives
$$|x|^k + |y|^k = 1.$$
This is the curve $C_p$ where $p = k$. The envelope of the family is the curve $C_p$ where $p = \dfrac{q}{r+1}$.
