# How do you find all the critical points to graph 9x^2-16y^2+18x+160y-247=0 including vertices, foci and asymptotes?

Oct 21, 2016

The center $\left(- 1 , 5\right)$
Vertices are $\left(- 1 , 2\right)$ and $\left(- 1 , 8\right)$
Foci: $\left(- 1 , 0\right)$ and $\left(- 1 , 10\right)$
Asymptotes:$y = \frac{3}{4} \left(x - - 1\right) + 5 \mathmr{and} y = - \frac{3}{4} \left(x - - 1\right) + 5$

#### Explanation:

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

Add $247 + 9 {h}^{2} - 16 {k}^{2}$ to both sides:

$9 {x}^{2} + 18 x + 9 {h}^{2} - 16 {y}^{2} + 160 y - 16 {k}^{2} = 247 + 9 {h}^{2} - 16 {k}^{2}$

Factor 9 and 16 from two quadratics on the left:

$9 \left({x}^{2} + 2 x + {h}^{2}\right) - 16 \left({y}^{2} - 10 y - {k}^{2}\right) = 247 + 9 {h}^{2} - 16 {k}^{2}$

Complete the squares, $h = - 1 , {h}^{2} = 1 , k = 5 , \mathmr{and} {k}^{2} = 25$

$9 {\left(x - - 1\right)}^{2} - 16 {\left(y - 5\right)}^{2} = 247 + 9 \left(1\right) - 16 \left(25\right)$

$9 {\left(x - - 1\right)}^{2} - 16 {\left(y - 5\right)}^{2} = - 144$

Divide both sides by -144:

${\left(y - 5\right)}^{2} / {3}^{2} - {\left(x - - 1\right)}^{2} / {4}^{2} = 1$

This is type 

The center, $\left(h , k\right)$ is at $\left(- 1 , 5\right)$

To find the vertices, let $x = - 1$

${\left(y - 5\right)}^{2} / {3}^{2} - {\left(- 1 - - 1\right)}^{2} / {4}^{2} = 1$

${\left(y - 5\right)}^{2} / {3}^{2} = 1$

${\left(y - 5\right)}^{2} = {3}^{2}$

$y - 5 = \pm {3}^{2}$

$y = 5 + 3 = 8$ and $y = 5 - 3 = 2$

Vertices are $\left(- 1 , 2\right)$ and $\left(- 1 , 8\right)$

The focal length, c,:

${c}^{2} = {a}^{2} + {b}^{2} = {3}^{2} + {4}^{2} = 25$

c = 5

$y = k - c = 5 - 5 = 0$ and y = k + c = 5 + 5 = 10

Foci: $\left(- 1 , 0\right)$ and $\left(- 1 , 10\right)$

Asymptotes:

$y - {y}_{0} = \pm \frac{a}{b} \left(x - {x}_{0}\right)$

$y = \frac{3}{4} \left(x - - 1\right) + 5 \mathmr{and} y = - \frac{3}{4} \left(x - - 1\right) + 5$