Post History

First of all, let's suppose that the points on the ellipse $(x_1,y_1),(x_2,y_2)$ are of the same distance from $(0,-2)$. We will get three equations from this assumption:

$$
\begin{cases}
x_1^2+3y_1^2=12 & (1)\\\\\\\\
x_2^2+3y_2^2=12 & (2)\\\\\\\\
x_1^2+(y_1+2)^2=x_2^2+(y_2+2)^2 & (3)
\end{cases}
$$

After substituting $x_1^2,x_2^2$ with $12-3y_1^2,12-3y_2^2$ in equation $(3)$ respectively, we will get $-2y_1^2+4y_1+16=-2y_2^2+4y_2+16$, which is equivalent to $(y_1-1)^2=(y_2-1)^2$. Now we will divide to two separate cases.

**Case 1:**

 If $(y_1-1),(y_2-1)$ are of the same sign (positive or negative), we can infer that $y_1=y_2$, and after equating $(1),(2)$ we will get $x_1^2+3y_1^2=x_2^2+3y_2^2$, Hence $x_1^2=x_2^2$. So there are two options again

**Subcase 1:** 

if $x_1=x_2$, then the two points are that same point. In this case the answer will depend on your definition of an equilateral triangle. if you consider a line as an equilateral triangle, then  every point on the ellipse will be a solution. If you will demand that the distance between all *three* (if we will consider the same point we got as two solutions) points will be the same, there is only one solution, that all the points will be $(0,-2)$. If you consider only triangles with three distinct points, we will have no solution in this case.

**Subcase 2:**

if $x_1=-x_2$, for simplicity i will denote $x_1=x\ y_1=y,\ x_2=-x,\ y_2=y$. The distance between the points $(x,y)$ and $(-x,y)$ is $2x$, and the distance between those two points to $(0,-2)$ is $\sqrt{x^2+(y+2)^2}$, so if we will require that this two length will be equal we will get the equation $4x^2=x^2+(y+2)^2$, and after substituting $x^2$ with $12-3y^2$ we will get 

$$
0=9y^2-36+y^2+4y+4=10y^2+4y-32=2(y+2)(5y-8)
$$ 

If $y=2$, then we will be at the last case, other wise we will have $y=\frac{8}{5}=1.6$, $|x|=\sqrt{12-3y^2}=\frac{6}{5}\sqrt{3}=1.2\sqrt{3}$, so the two points will be $(\pm 1.2\sqrt{3},1.6)$ as in the book.

**Case 2:**

If the signs of $(y_1-1),(y_2-1)$ are opposite, we will get $y_1=2-y_2$. From equations $(1),(2)$ we will get $x_1^2-x_2^2=3(y_2^2-y_1^2)=3(y_2+y_1)(y_2-y_1)=12(y_2-1)=12(1-y_1)$. Now, if we will require that the distance between $(x_1,y_1),(x_2,y_2)$ equals to the distance between $(x_1,y_1),(0,-2)$ we could write

$$
\begin{align}
(x_1-x_2)^2+(y_1-y_2)^2 & =x_1^2+(y_1+2)^2\\\\
\\\\
x_1^2+x_2^2-2x_1x_2+(2y_1-2)^2 & =x_1^2+y_1^2+4y_1+4\\\\
\\\\
x_2^2-2x_1x_2+4y_1^2-8y_1+4 & =y_1^2+4y_1+4\\\\
\\\\
-2x_1x_2+x_2^2+3y_1^2-12y_1 & =0\ \/  {\color{gray}\text{because}\ 3y_1^2=12-x_1^2\ \text{we will get} }\\\\
\\\\
-2x_1x_2+12+x_2^2-x_1^2-12y_1 & =0\ \/  {\color{gray}\text{remember that}\ x_1^2-x_2^2=12(1-y_1)}\\\\
\\\\
-2x_1x_2-12(1-y_1)+12(1-y_1) & =0\\\\
\\\\
-2x_1x_2 & =0
\end{align}
$$ 

Thus $x_1$ or $x_2$ are vanishing. Without loss of generality i will assume that $x_1=0$.  From that we will get that $y_1=\pm 2$. Because $y_1+y_2=2$ and $y_2\leq 2$, we will conclude that $y_1\geq 0$ so $y_1=2$. Because $y_1+y_2=2$ we will get that $y_2=0$, so $x_2=\pm\sqrt{12}$, as you got as well.

**Conclusion:** All the solutions are $\left\[(1.2\sqrt{3},1.6),(-1.2\sqrt{3},1.6)\right\],\ \left\[(0,2),(\sqrt{12},0)\right\],\ \left[(0,2),(-\sqrt{12},0)\right]$.