Let's write the equation of the curve for which the sum of the distance from the point $(-2,0)$ and twice the distance from the point $(1,0)$ is constant.

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Let the point $(-2,0)$ be $F_{1}$, the point $(1,0)$ be $F_{2}$, and the arbitrary point on the curve be $P(x, y)$, with $P F_{1}=r_{1}, P F_{2}=r_{2}$.

In the triangle $P F_{1} F_{2}$: $P F_{1}+P F_{2} \geq F_{1} F_{2}=3$, or $r_{1}+r_{2} \geq 3$, thus $r_{1}+2 r_{2} \geq 3+r_{2}$;

furthermore:

$$
\begin{array}{ll}
r_{1}-r_{2}<3 \text{ and } r_{2}-r_{1}<3 \text{, so } r_{1}-2 r_{2}<3-r_{2} \text{ and } 2 r_{2}-r_{1}<3-r_{2} \\
\text{If } r_{1}-2 r_{2} \geq 3-r_{2} \text{ or } 2 r_{2}-r_{1} \geq 3-r_{2}, \text{ then } r_{1}-2 r_{2}-a=0 \text{ or } 2 r_{2}-r_{1}-a=0, \text{ only if } a<3-r_{2} \text{.} \\
\text{If } r_{1}+2 r_{2} \geq 3+r_{2}, \text{ then } r_{1}+2 r_{2}=a, \text{ or } r_{1}+2 r_{2}-a=0. \text{ Conversely: if the product is 0 and } a>3+r_{2}, \text{ then only } r_{1}+2 r_{2}-a=0, \text{ or } P \text{ is on the desired curve.}
\end{array}
$$

The equation $\left(r_{1}+2 r_{2}-a\right)\left(r_{1}-2 r_{2}-a\right)\left(2 r_{2}-r_{1}-a\right)=0$ is thus the equation of the desired curve, if $a>3+r_{2}$. The equation can be made more symmetric by multiplying by $\left(r_{1}+2 r_{2}+a\right)$, which cannot be 0:

$$
\left(r_{1}+2 r_{2}-a\right)\left(r_{1}+2 r_{2}+a\right)\left(r_{1}-2 r_{2}-a\right)\left(2 r_{2}-r_{1}-a\right)=0
$$

Applying the identity for the product of the sum and difference of two terms:

$$
\begin{gathered}
{\left[\left(r_{1}+2 r_{2}\right)^{2}-a^{2}\right]\left[a^{2}-\left(r_{1}-2 r_{2}\right)^{2}\right]=0} \\
\left(r_{1}^{2}+4 r_{1} r_{2}+4 r_{2}^{2}-a^{2}\right)\left(a^{2}-r_{1}^{2}+4 r_{1} r_{2}-4 r_{2}^{2}\right)=0 \\
16 r_{1}^{2} r_{2}^{2}-\left(r_{1}^{2}+4 r_{2}^{2}-a^{2}\right)^{2}=0 \\
16 r_{1}^{2} r_{2}^{2}-\left(r_{1}^{4}+8 r_{1}^{2} r_{2}^{2}+16 r_{2}^{4}-2 r_{1}^{2} a^{2}-8 r_{2}^{2} a^{2}+a^{4}\right)=0 \\
-r_{1}^{4}+8 r_{1}^{2} r_{2}^{2}-16 r_{2}^{4}+2 a^{2}\left(r_{1}^{2}+4 r_{2}^{2}\right)-a^{4}=0 \\
-\left(r_{1}^{2}-4 r_{2}^{2}\right)^{2}+2 a^{2}\left(r_{1}^{2}+4 r_{2}^{2}\right)-a^{4}=0 \\
r_{1}^{2}=P F_{1}^{2}=(x+2)^{2}+y^{2} \quad \text { and } \quad r_{2}^{2}=P F_{2}^{2}=(x-1)^{2}+y^{2}
\end{gathered}
$$

Thus

$$
\begin{gathered}
r_{1}^{2}-4 r_{2}^{2}=x^{2}+4 x+4+y^{2}-4 x^{2}+8 x-4-4 y^{2}= \\
=-3\left(x^{2}+y^{2}-4 x\right)
\end{gathered}
$$

and

$$
\begin{gathered}
r_{1}^{2}+4 r_{2}^{2}=x^{2}+4 x+4+y^{2}+4 x^{2}-8 x+4+4 y^{2}= \\
=5 x^{2}+5 y^{2}-4 x+8
\end{gathered}
$$

Substituting these:

$$
\begin{gathered}
-9\left(x^{2}+y^{2}-4 x\right)^{2}+2 a^{2}\left(5 x^{2}+5 y^{2}-4 x+8\right)-a^{4}=0 \\
9\left[\left(x^{2}+y^{2}\right)^{2}-8 x\left(x^{2}+y^{2}\right)+16 x^{2}\right]-2 a^{2}\left(5 x^{2}+5 y^{2}-4 x+8\right)+a^{4}=0 \\
9\left(x^{2}+y^{2}\right)^{2}-\left(x^{2}+y^{2}\right)\left(72 x+10 a^{2}\right)+144 x^{2}+8 a^{2} x-16 a^{2}+a^{4}=0 \\
9\left(x^{2}+y^{2}\right)^{2}-\left(x^{2}+y^{2}\right)\left(72 x+10 a^{2}\right)+\left(4 x+a^{2}\right)^{2}+16\left(8 x^{2}-a^{2}\right)=0
\end{gathered}
$$