How do you find all the critical points to graph x^2 - y^2 - 10x - 10y - 1= 0x2y210x10y1=0 including vertices, foci and asymptotes?

1 Answer
Oct 20, 2016

The vertices are (6,-5)(6,5) and (4,-5)(4,5)
The foci are (5+sqrt2,-5(5+2,5 and (5-sqrt2,-5)(52,5)
The asymptotes are y=x-10y=x10 and y graph{x^2-y^2-10*x-10*y-1=0 [-13.83, 26.17, -15.28, 4.72]} =-xygraph{x2y210x10y1=0[13.83,26.17,15.28,4.72]}=x

Explanation:

(x^2-10x)-(y^2+10y)=1(x210x)(y2+10y)=1
Completing the squares (x-5)^2-(y+5)^5=1(x5)2(y+5)5=1
((x-5)^2)/1-((y+5)^2)/1=1(x5)21(y+5)21=1
So the certer is (5,-5)(5,5)
The vertices are (6,-5)(6,5) and (4,-5)(4,5)
The slope of the asymptotes are 11 and -11
The equations of the asymptotes are y=-5+1*(x-5)=x-10y=5+1(x5)=x10 and y=-5-1*(x-5)=-xy=51(x5)=x
we need to calculate c=+-sqrt(1+1)=sqrt2c=±1+1=2
c is the distance between the certer and the focus
And the foci are (5+sqrt2,-5(5+2,5 and (5-sqrt2,-5)(52,5)