# How do you identify the vertices, foci, and direction of (y-1)^2/9-(x+1)^2/16=1?

Nov 10, 2016

The vertices are $= \left(- 1 , 4\right)$ and $= \left(- 1 , - 2\right)$
The foci are F$= \left(- 1 , 6\right)$ and F'$= \left(- 1 , - 4\right)$
The equations of the asymptotes are $y = 1 + \frac{3}{4} \left(x + 1\right)$ and $y = 1 - \frac{3}{4} \left(x + 1\right)$

#### Explanation:

This is the equation of an updown hyperbola.
The standard equation is ${\left(y - k\right)}^{2} / {b}^{2} - {\left(x - h\right)}^{2} / {a}^{2} = 1$
The center is $\left(h , k\right) = \left(- 1 , 1\right)$

The vertices are $\left(h , k + b\right) = \left(- 1 , 4\right)$ and $\left(h , k - b\right) = \left(- 1 , - 2\right)$

The slopes of the asymptotes are $\pm \frac{b}{a} = \pm \frac{3}{4}$
The equations of the asymptotes are $y = k \pm \frac{b}{a} \left(x - h\right)$
$y = 1 + \frac{3}{4} \left(x + 1\right)$ and $y = 1 - \frac{3}{4} \left(x + 1\right)$

To calculate the foci, we need $c = \pm \sqrt{{a}^{2} + {b}^{2}}$
So $c = \pm \sqrt{9 + 16} = \pm \sqrt{25} = \pm 5$
The foci are $= \left(h , k \pm c\right)$
so F$= \left(- 1 , 6\right)$ and F' $\left(- 1 , - 4\right)$
graph{((y-1)^2/9-(x+1)^2/16-1)(y-1-3/4(x+1))(y-1+3/4(x+1))=0 [-22.82, 22.77, -11.4, 11.4]}