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

Jul 30, 2018

Please see the explanation below

Explanation:

The equation of a hyperbola with a horizontal transverse axis is

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

Our equation is

${\left(x + 2\right)}^{2} / 4 - {\left(y - 5\right)}^{2} / 25 = 1$

$a = 2$

$b = 5$

$h = - 2$

$k = 5$

$c = \sqrt{{b}^{2} + {c}^{2}} = \sqrt{25 + 4} = \sqrt{29}$

The center is $C = \left(h , k\right) = \left(- 2 , 5\right)$

The vertices are $A = \left(h + a , k\right) = \left(0 , 5\right)$

and

$C ' = \left(h - a , k\right) = \left(- 4 , 5\right)$

The foci are $F = \left(h + c , k\right) = \left(- 2 + \sqrt{29} , 5\right)$

and

$F ' = \left(- 2 - \sqrt{29} , 5\right)$

The asymptotes are

${\left(x + 2\right)}^{2} / 4 - {\left(y - 5\right)}^{2} / 25 = 0$

${\left(x + 2\right)}^{2} / 4 = {\left(y - 5\right)}^{2} / 25$

$\frac{y - 5}{5} = \pm \frac{x + 2}{2}$

graph{((x+2)^2/4-(y-5)^2/25-1)((y-5)/5-(x+2)/2)((y-5)/5+(x+2)/2)=0 [-12.25, 7.75, 0.2, 10.2]}