# How do you find all the critical points to graph x^2 - 9y^2 + 2x - 54y + 80 = 0 including vertices, foci and asympotes?

Oct 19, 2016

Please see the explanation.

#### Explanation:

Add ${h}^{2} - 9 {k}^{2} - 80$ to both sides:

${x}^{2} + 2 x + {h}^{2} - 9 {y}^{2} - 54 y - 9 {k}^{2} = {h}^{2} - 9 {k}^{2} - 80$

Set the right side of the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ equal to the terms ${x}^{2} + 2 x + {h}^{2}$:

${x}^{2} - 2 h x + {h}^{2} = {x}^{2} + 2 x + {h}^{2}$

Solve for $h$ and ${h}^{2}$:

$- 2 h x + {h}^{2} = 2 x + {h}^{2}$

$- 2 h x = 2 x$

$h = - 1$ and ${h}^{2} = 1$

Substitute the left side of the pattern (along with the value of h) for the 3 terms on the left and substitute 1 for ${h}^{2}$ on the right:

${\left(x - - 1\right)}^{2} - 9 {y}^{2} - 54 y - 9 {k}^{2} = 1 - 9 {k}^{2} - 80$

Factor -9 from the y terms:

${\left(x - 1\right)}^{2} - 9 \left({y}^{2} + 6 y + {k}^{2}\right) = 1 - 9 {k}^{2} - 80$

Set the right side of the pattern ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$ equal to the terms ${y}^{2} + 6 y + {k}^{2}$:

${y}^{2} - 2 k y + {k}^{2} = {y}^{2} + 6 y + {k}^{2}$

Solve for $k$ and ${k}^{2}$:

$- 2 k y = 6 y$

$k = - 3$ and ${k}^{2} = 9$

Substitute the left side on the pattern (along with the value of k) into the ()s on the left and 9 for ${k}^{2}$ on the right:

${\left(x - - 1\right)}^{2} - 9 \left({\left(y - - 3\right)}^{2}\right) = 1 - 9 \left(9\right) - 80$

Simplify the right side:

${\left(x - - 1\right)}^{2} - 9 \left({\left(y - - 3\right)}^{2}\right) = - 160$

Divide both sides by -160:

$9 \frac{{\left(y - - 3\right)}^{2}}{160} - {\left(x - - 1\right)}^{2} / 160 = 1$

Write denominators as squares:

${\left(y - - 3\right)}^{2} / {\left(\frac{4 \sqrt{10}}{3}\right)}^{2} - {\left(x - - 1\right)}^{2} / {\left(4 \sqrt{10}\right)}^{2} = 1$

Center:$\left(- 1 , - 3\right)$
To find the vertices, make the negative term disappear by setting $x = - 1$:

${\left(y - - 3\right)}^{2} / {\left(\frac{4 \sqrt{10}}{3}\right)}^{2} = 1$

${\left(y - - 3\right)}^{2} = {\left(\frac{4 \sqrt{10}}{3}\right)}^{2}$

$y - - 3 = \frac{4 \sqrt{10}}{3}$ and $y - - 3 = - \frac{4 \sqrt{10}}{3}$

$y = - 3 + \frac{4 \sqrt{10}}{3}$ and $y = - 3 - \frac{4 \sqrt{10}}{3}$
Vertices:$\left(- 1 , - 3 + \frac{4 \sqrt{10}}{3}\right)$ and $\left(- 1 , - 3 - \frac{4 \sqrt{10}}{3}\right)$

To find the focal distance, c, take the square root of the sum of the squares of the denominators:

$c = \sqrt{\frac{160}{9} + 160}$

$c = \sqrt{\frac{160}{9} + 160}$

$c = \frac{40}{3}$

The foci are at : $\left(- 1 , - 3 + \frac{40}{3}\right)$ and $\left(- 1 , - 3 - \frac{40}{3}\right)$

Foci:$\left(- 1 , \frac{31}{3}\right)$ and $\left(- 1 , - \frac{49}{3}\right)$

To find the asymptotes, please observe that dividing the first denominator by the second gives the slopes of $\frac{1}{3}$ and $- \frac{1}{3}$. All that is left to do is to force the lines with those two slopes, through the center point.

$y = \frac{1}{3} \left(x - - 1\right) - 3$
$y = - \frac{1}{3} \left(x - - 1\right) - 3$