# How do you find all the critical points to graph (x+3)^2/25-(y+5)^2/4=1 including vertices, foci and asymptotes?

##### 1 Answer
Oct 29, 2016

The vertices are $\left(2 , - 5\right)$ and $\left(- 8 , - 5\right)$
The equations of the asymptotes are $y = - 5 + \frac{2}{5} \left(x + 3\right)$ and $y = - 5 - \frac{2}{5} \left(x + 3\right)$
The foci are $\left(- 3 + \sqrt{29} , - 5\right)$ and $\left(- 3 - \sqrt{29} , - 5\right)$

#### Explanation:

Our equation is ${\left(x + 3\right)}^{2} / 25 - {\left(y + 5\right)}^{2} / 4 = 1$
This is the equation of a left to right hyperbola
The typical equation of a hyperbola is ${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$
The center is $\left(h , k\right) = \left(- 3 , - 5\right)$
The vertices are $\left(h + a , k\right) = \left(- 3 + 5 , - 5\right) = \left(2 , - 5\right)$ and $\left(h - a , k\right) = \left(- 3 - 5 , - 5\right) = \left(- 8 , - 5\right)$
The slope of the asymptotes are $\frac{b}{a} = \frac{2}{5}$ and $- \frac{b}{a} = - \frac{2}{5}$

The equation of the asymptotes are $y = k + \frac{b}{a} \left(x - h\right)$ and $y - \frac{b}{a} \left(x - h\right)$

So the equations are $y = - 5 + \frac{2}{5} \left(x + 3\right)$ and $y = - 5 - \frac{2}{5} \left(x + 3\right)$

In order to calculate the foci, we need $c = \pm \sqrt{{a}^{2} + {b}^{2}} = \pm \sqrt{25 + 4} = \pm \sqrt{29}$
And the foci are $\left(- 3 + \sqrt{29} , - 5\right)$ and $\left(- 3 - \sqrt{29} , - 5\right)$

graph{(x+3)^2/25-(y+5)^2/4=1 [-20.91, 18.56, -13.78, 5.97]}