Question

Find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation by hand. Verify your graph using a graphing utility.

$x^2-y^2-2x-2y-1=0$

Short answer

The center of the hyperbola is $\left(1,-1\right)$, foci are$\left(1\pm \sqrt{2},-1\right)$, vertices are$\left(2,-1\right),\left(0,-1\right)$, asymptotes are$y+1=\pm \left(x-1\right)$and transverse is parallel to$x$-axis.

The graph of the hyperbola is

Step-by-step solution

Given data

The hyperbola is given by

$x^2-y^2-2x-2y-1=0$

Concept used

The hyperbola with center $\left(h,k\right)$and transverse axis parallel to $x$-axis is given by,

role="math" localid="1647146663576" $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$; $b^2=c^2-a^2$

where foci are $\left(h\pm c,k\right)$, vertices are$\left(h\pm a,k\right)$and asymptotes are$y-k=\pm \frac{b}{a}\left(x-h\right)$.

Application 

The given hyperbola is

$x^2-y^2-2x-2y-1=0$

Convert into standard form of hyperbolas.

Grouping terms,

role="math" localid="1647147350929" $\left(x^2-2x\right)-\left(y^2+2y\right)=1$

Completing each square,

$\left(x^2-2x+1\right)-\left(y^2+2y+1\right)=1$

${\left(x-1\right)}^2-{\left(y+1\right)}^2=1$

This is the standard form the hyperbolas.

So, $a=1,b=1$

$h=1,k=-1$

Since $c^2=a^2+b^2$,

$c^2=1+1$

role="math" $c^2=2$

$c=\sqrt{2}$

The center of the hyperbola is $\left(h,k\right)\equiv \left(1,-1\right)$,

The foci are $\left(h\pm c,k\right)\equiv \left(1\pm \sqrt{2},-1\right)$,

The vertices are role="math" localid="1647147717332" $\left(h+a,k\right)\equiv \left(1+1,-1\right)\equiv \left(2,-1\right)$and

$\left(h-a,k\right)\equiv \left(1-1,-1\right)\equiv \left(0,-1\right)$

Since foci and vertices all line on the line $y=-1$, the transverse is parallel to $x$-axis.

The asymptotes of the hyperbola is

$y+1=\pm \frac{1}{1}\left(x-1\right)$

role="math" localid="1647148342944" $y+1=\pm \left(x-1\right)$

Graph

The graph of the hyperbola is