# How do you find all the critical points to graph x^2 - y^2 - 10x - 10y - 1= 0 including vertices, foci and asymptotes?

Oct 20, 2016

The vertices are $\left(6 , - 5\right)$ and $\left(4 , - 5\right)$
The foci are (5+sqrt2,-5 and $\left(5 - \sqrt{2} , - 5\right)$
The asymptotes are $y = x - 10$ and $y g r a p h \left\{{x}^{2} - {y}^{2} - 10 \cdot x - 10 \cdot y - 1 = 0 \left[- 13.83 , 26.17 , - 15.28 , 4.72\right]\right\} = - x$

#### Explanation:

$\left({x}^{2} - 10 x\right) - \left({y}^{2} + 10 y\right) = 1$
Completing the squares ${\left(x - 5\right)}^{2} - {\left(y + 5\right)}^{5} = 1$
$\frac{{\left(x - 5\right)}^{2}}{1} - \frac{{\left(y + 5\right)}^{2}}{1} = 1$
So the certer is $\left(5 , - 5\right)$
The vertices are $\left(6 , - 5\right)$ and $\left(4 , - 5\right)$
The slope of the asymptotes are $1$ and $- 1$
The equations of the asymptotes are $y = - 5 + 1 \cdot \left(x - 5\right) = x - 10$ and $y = - 5 - 1 \cdot \left(x - 5\right) = - x$
we need to calculate $c = \pm \sqrt{1 + 1} = \sqrt{2}$
c is the distance between the certer and the focus
And the foci are (5+sqrt2,-5 and $\left(5 - \sqrt{2} , - 5\right)$