Parabolic Representation The points $(-1/2, 0)$, $\left(0, \frac{\sqrt3}{2}\right)$, $(1/2, 0)$ are the vertices of an equilateral triangle. A parabolic equation that contains the three points is 
$$y=
(-2 \sqrt3)x^2+\frac{\sqrt{3}}{2} \,.$$ How can the other two parabolas, that are congruent to the first parabola and contain the same three points, be represented? Can this be done with parametric equations or do I need a higher level of understanding?   

Since you are given the three points, there is a straight vectorial approach that you can follow.
Premised that
the parabola $y=x^2$,  the parabola  with vertex in $O=(0,0)$ and passing through
 $U_{-1}=(-1,1)$ and  $U_{1}=(1,1)$, can be written as:
$$
\mathop {OP}\limits^ \to   \cdot \mathbf{u}_y  = \left( {\mathop {OP}\limits^ \to   \cdot \mathbf{u}_x } \right)^2 
$$
where $\mathop {OP}\limits^ \to   = (x,y)$ is the position vector, and
$\mathbf{u}_x  = \left( {1,0} \right)\quad \mathbf{u}_y  = \left( {0,1} \right)$
are the unit vectors on the axes, and that we have
$$
\mathop {OU_1 }\limits^ \to   = \mathbf{u}_y  + \mathbf{u}_x \quad \mathop {OU_{ - 1} }\limits^ \to   = \mathbf{u}_y  - \mathbf{u}_x 
$$
then
in the case of the three symmetric points under consideration
$$
A = \left( { - 1/2,\;0} \right)\quad B = \left( {1/2,\;0} \right)\quad C = \left( {0,\;\sqrt 3 /2} \right)
$$
let's consider the vectors 
$$
\begin{gathered}
  \mathbf{v} = \frac{1}
{2}\left( {\mathop {AB}\limits^ \to   - \mathop {AC}\limits^ \to  } \right) = \frac{1}
{2}\left( {\left( {x_B  - x_C } \right),\;\left( {y_B  - y_C } \right)} \right) \hfill \\
  \mathbf{w} = \frac{1}
{2}\left( {\mathop {AB}\limits^ \to   + \mathop {AC}\limits^ \to  } \right) = \frac{1}
{2}\left( {\left( {x_B  + x_C  - 2x_A } \right),\;\left( {y_B  + y_C  - 2y_A } \right)} \right) \hfill \\ 
\end{gathered} 
$$
as shown in the figure

we want that
$$
\left\{ \begin{gathered}
  \mathop {AB}\limits^ \to   \cdot \mathbf{w} = \left\| \mathbf{w} \right\|^{\,2} \quad  \Rightarrow \quad 1 \hfill \\
  \mathop {AB}\limits^ \to   \cdot \mathbf{v} = \left\| \mathbf{v} \right\|^{\,2} \quad  \Rightarrow \quad 1 \hfill \\ 
\end{gathered}  \right.
$$
therefore the parabola with vertex in $A$ e passing through $B$ and $C$ can be written as:
$$
\frac{{\mathop {AP}\limits^ \to   \cdot \mathbf{w}}}
{{\left\| \mathbf{w} \right\|^{\,2} }} = \left( {\frac{{\mathop {AP}\limits^ \to   \cdot \mathbf{v}}}
{{\left\| \mathbf{v} \right\|^{\,2} }}} \right)^2 
$$
i.e.
$$
\left( {\left( {x + 1/2} \right),\;y} \right) \cdot \frac{1}
{2}\left( {\begin{array}{*{20}c}
   {3/2}  \\
   {\sqrt 3 /2}  \\
 \end{array} } \right)\frac{4}
{3} = \left( {\left( {\left( {x + 1/2} \right),\;y} \right) \cdot \frac{1}
{2}\left( {\begin{array}{*{20}c}
   {1/2}  \\
   { - \sqrt 3 /2}  \\
 \end{array} } \right)4} \right)^2 
$$
thus
$$
2\left( {6x + 3 + 2\sqrt 3 y} \right) = 3\left( {2x + 1 - 2\sqrt 3 y} \right)^2 
$$
or
$$
12x^2  + 36y^2  - 24\sqrt 3 xy - 16\sqrt 3 y - 3 = 0
$$
