# How do you identify the vertices, foci, and direction of -10y-y^2=-4x^2-72x-199?

Mar 15, 2017

Write the equation in one of the two standard forms then use information to find the vertices and foci.

#### Explanation:

Given:

$- 10 y - {y}^{2} = - 4 {x}^{2} - 72 x - 199$

Add $4 {x}^{2} + 72 x - {k}^{2} + 4 {h}^{2}$ to both sides of the equation:

$4 {x}^{2} + 72 x + 4 {h}^{2} - {y}^{2} - 10 y - {k}^{2} = - 199 + 4 {h}^{2} - {k}^{2}$

Remove a common factor of 4 from the first 3 terms and a common factor of -1 from the next 3 terms:

$4 \left({x}^{2} + 18 x + {h}^{2}\right) - \left({y}^{2} + 10 y + {k}^{2}\right) = - 199 + 4 {h}^{2} - {k}^{2} \text{ [1]}$

Use the right side of the patterns ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ and ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$ and the terms in the parenthesis in equation [1] to find the values of h and k:

${x}^{2} - 2 h x + {h}^{2} = {x}^{2} + 18 x + {h}^{2}$ and ${y}^{2} - 2 k y + {k}^{2} = {y}^{2} + 10 y + {k}^{2}$

$- 2 h x = 18 x$ and $- 2 k y = 10 y$

$h = - 9$ and $k = - 5$

Substitute the left side of the patterns into equation [1]:

$4 {\left(x - h\right)}^{2} - {\left(y - k\right)}^{2} = - 199 + 4 {h}^{2} - {k}^{2} \text{ [2]}$

Substitute the values for h and k into equation [2]:

$4 {\left(x - - 9\right)}^{2} - {\left(y - - 5\right)}^{2} = - 199 + 4 {\left(- 9\right)}^{2} - {\left(- 5\right)}^{2} \text{ [3]}$

Simplify the right side of equation [3]:

$4 {\left(x - - 9\right)}^{2} - {\left(y - - 5\right)}^{2} = 100 \text{ [4]}$

Divide both sides of the equation by 100:

${\left(x - - 9\right)}^{2} / 25 - {\left(y - - 5\right)}^{2} / 100 = 1 \text{ [5]}$

Write the denominators as squares:

${\left(x - - 9\right)}^{2} / {5}^{2} - {\left(y - - 5\right)}^{2} / {10}^{2} = 1 \text{ [6]}$

Equation [6] is the standard form for a hyperbola with a horizontal transverse axis.

Referring to the standard form, the vertices are at $\left(h - a , k\right) \mathmr{and} \left(h + a , k\right)$

Substituting $a = 5 , h = - 9 , \mathmr{and} k = - 5$

$\left(- 9 - 5 , - 5\right) \mathmr{and} \left(- 9 + 5 , - 6\right)$

The vertices are $\left(- 16 , - 5\right) \mathmr{and} \left(- 4 , - 5\right)$

Referring to the standard form, the foci are at $\left(h - c , k\right) \mathmr{and} \left(h + c , k\right)$

Use $c = \sqrt{{a}^{2} + {b}^{2}}$ with a = 5 and b = 10:

$c = \sqrt{125}$

The foci are $\left(- 9 - \sqrt{125} , - 5\right) \mathmr{and} \left(- 9 + \sqrt{125} , - 5\right)$

Referring to the standard form, the asymptotes are at:

$y = - \frac{b}{a} \left(x - h\right) + k \mathmr{and} y = \frac{b}{a} \left(x - h\right) + k$

Substituting $b = 10 , a = 5 , h = - 9 , \mathmr{and} k = - 5$

The asymptotes are: $y = - 2 \left(x - - 9\right) - 5 \mathmr{and} y = 2 \left(x - - 9\right) - 5$