# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola (y-4)^2/16-(x+2)^2/9=1?

Given equation of hyperbola:

$\setminus \frac{{\left(y - 4\right)}^{2}}{{4}^{2}} - \setminus \frac{{\left(x + 2\right)}^{2}}{{3}^{2}} = 1$

Comparing above equation with standard form of vertical hyperbola: $\setminus \frac{{Y}^{2}}{{a}^{2}} - \setminus \frac{{X}^{2}}{{b}^{2}} = 1$ we get

$X = x + 2 , Y = y - 4 , a = 4 , b = 3$

$\setminus \textrm{e \mathcal{e} n t r i c i t y} , e = \setminus \sqrt{1 + \setminus \frac{{b}^{2}}{{a}^{2}}} = \setminus \sqrt{1 + \setminus \frac{{3}^{2}}{{4}^{2}}} = \frac{5}{4}$

Center of hyperbola:

$X = 0 , Y = 0$

$x + 2 = 0 , y - 4 = 0$

$x = - 2 , y = 4$

$\setminus \textrm{C e n t e r} \setminus \equiv \left(- 2 , 4\right)$

Now, the vertices of hyper bola

$X = 0 , Y = \setminus \pm a$

$x + 2 = 0 , y - 4 = \setminus \pm 4$

$x = - 2 , y = 4 \setminus \pm 4$

\text {Vertices} (-2, 8)\ & \ (-2, 0)

Now, the focii of hyper bola

$X = 0 , Y = \setminus \pm a e$

$x + 2 = 0 , y - 4 = \setminus \pm 4 \setminus \cdot \frac{5}{4}$

$x = - 2 , y = 4 \setminus \pm 5$

\text {focii}, (-2, 9)\ &\ (-2, -1)

Asymptotes of hyperbola

$Y = \setminus \pm \frac{a}{b} X$

$y - 4 = \setminus \pm \frac{4}{3} \left(x + 2\right)$

$y = \setminus \pm \frac{4}{3} \left(x + 2\right) + 4$