How do you find all the critical points to graph x^2 - y^2 - 10x - 10y - 1= 0x2−y2−10x−10y−1=0 including vertices, foci and asymptotes?

1 Answer
Narad T.
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)(5−√2,−5)
The asymptotes are y=x-10y=x−10 and y graph{x^2-y^2-10*x-10*y-1=0 [-13.83, 26.17, -15.28, 4.72]} =-xygraph{x2−y2−10⋅x−10⋅y−1=0[−13.83,26.17,−15.28,4.72]}=−x

Explanation:

(x^2-10x)-(y^2+10y)=1(x2−10x)−(y2+10y)=1
Completing the squares (x-5)^2-(y+5)^5=1(x−5)2−(y+5)5=1
((x-5)^2)/1-((y+5)^2)/1=1(x−5)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 -1−1
The equations of the asymptotes are y=-5+1*(x-5)=x-10y=−5+1⋅(x−5)=x−10 and y=-5-1*(x-5)=-xy=−5−1⋅(x−5)=−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)(5−√2,−5)

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