# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola y^2-3x^2+6y+6x-18=0?

#### Explanation:

Given equation of hyperbola:

${y}^{2} - 3 {x}^{2} + 6 y + 6 x - 18 = 0$

$\left({y}^{2} + 6 y + 9\right) - 3 \left({x}^{2} - 2 x + 1\right) - 9 + 3 - 18 = 0$

${\left(y + 3\right)}^{2} - 3 {\left(x - 1\right)}^{2} = 24$

$\setminus \frac{{\left(y + 3\right)}^{2}}{24} - \setminus \frac{3 {\left(x - 1\right)}^{2}}{24} = 1$

$\setminus \frac{{\left(y + 3\right)}^{2}}{{\left(2 \setminus \sqrt{6}\right)}^{2}} - \setminus \frac{{\left(x - 1\right)}^{2}}{{\left(2 \setminus \sqrt{2}\right)}^{2}} = 1$

The above vertical hyperbola: ${y}^{2} / {a}^{2} - {x}^{2} / {b}^{2} = 1$ has

Center: $\left(x - 1 = 0 , y + 3 = 0\right) \setminus \equiv \left(1 , - 3\right)$

Eccentricity: $e = \setminus \sqrt{1 + {b}^{2} / {a}^{2}} = \setminus \sqrt{1 + \frac{8}{24}} = \frac{2}{\setminus} \sqrt{3}$

The vertices: $\left(x - 1 = 0 , y + 3 = \setminus \pm a\right) \setminus \equiv \left(1 , - 3 \setminus \pm 2 \setminus \sqrt{6}\right)$ &

$\left(x - 1 = \setminus \pm b , y + 3 = 0\right) \setminus \equiv \left(1 \setminus \pm 2 \setminus \sqrt{2} , - 3\right)$

Focii: $\left(x - 1 = 0 , y + 3 = \setminus \pm a e\right) \setminus \equiv \left(1 , - 3 \setminus \pm 4 \setminus \sqrt{2}\right)$

Asymptotes: $y = \setminus \pm \frac{a}{b} x$

$y + 3 = \setminus \pm \left(\frac{2 \setminus \sqrt{6}}{2 \setminus \sqrt{2}}\right) \left(x - 1\right)$

$y = - 3 \setminus \pm \setminus \sqrt{3} \left(x - 1\right)$