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

1 Answer
Oct 20, 2016

Answer:

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)#