Question

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

Answer 1

The vertices are `(6,-5)` and `(4,-5)`
The foci are `(5+sqrt2,-5` and `(5-sqrt2,-5)`
The asymptotes are `y=x-10` and `y graph{x^2-y^2-10*x-10*y-1=0 [-13.83, 26.17, -15.28, 4.72]} =-x`

Explanation:

`(x^2-10x)-(y^2+10y)=1`
Completing the squares `(x-5)^2-(y+5)^5=1`
`((x-5)^2)/1-((y+5)^2)/1=1`
So the certer is `(5,-5)`
The vertices are `(6,-5)` and `(4,-5)`
The slope of the asymptotes are `1` and `-1`
The equations of the asymptotes are `y=-5+1*(x-5)=x-10` and `y=-5-1*(x-5)=-x`
we need to calculate `c=+-sqrt(1+1)=sqrt2`
c is the distance between the certer and the focus
And the foci are `(5+sqrt2,-5` and `(5-sqrt2,-5)`