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

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

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Answer 1 · 2016-10-20T04:49:30

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

Explanation:

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

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

1 Answer

Explanation:

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

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Explanation:

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