$\frac{x^2}{16}-\frac{(y+1)^2}{9}=1$

$\text{Asymptotes: }y=\pm\frac{4}{3}x-1$

Center: (0, −1)
Vertices: (± 4, − 1)

$x=-\frac{1}{3}(y-1)^2+2$