Answer
Hyperbola
Lines of symmetry: $x=0,y=0$
Domain : $R^2$
Range: $[-\infty,5/\sqrt3] \cup [5/\sqrt3,\infty]$
Work Step by Step
The equation $3(y-0)^2/25-(x-0)^2/25=1$ is a Hyperbola.
The Lines of symmetry are $x=0,y=0$.
Domain is $R^2$.
By solving the equation for $y$ we get $$\begin{aligned}
3y^2&=x^2+25\\
y^2&=\frac{x^2+25}{3}\\
y&=\pm\sqrt{\frac{x^2+25}{3}}
\end{aligned}$$ Range is $x=[-\infty,5/\sqrt3] \cup [5/\sqrt3,\infty]$.
