Question 329720
Please sketch this hyperbola and show me how you did it: x^2-y^2=1.
<pre><font size = 4 color = "indigo"><b>
The equation

{{{x^2/a^2-y^2/b^2=1}}} is the equation of a hyperbola that opens
right and left with these properties:

1.  The center is at the origin (0,0).

2.  The vertices are the ends of the transverse axis (-a,0) and (a,0).

3.  The endpoints of the conjugate axis are (0,b) and (0,-b).

4.  The foci are the points (-c,0) and (c,0) where {{{c=sqrt(a^2+b^2)}}}

5.  The defining rectangle has the 4 corners 
    (-a,b), (-a,-b), (a,-b), and (a,b)

6.  The asymptotes are the extended diagonals of the defining rectangle
    and have equations {{{y}}}{{{""=""}}}{{{"" +- (b/a)}}}{{{x}}}

Your equation is 

{{{x^2-y^2=1}}}

Write it as 

{{{x^2/1-y^2/1=1}}}

and we see that {{{a^2=1}}} and {{{b^2=1}}}, so {{{a=1}}} and {{{b=1}}},

so:

1.  The center is at the origin (0,0).

2.  The vertices are the ends of the transverse axis (-1,0) and (1,0).

3.  The endpoints of the conjugate axis are (0,1) and (0,-1).

4.  The foci are the points (-c,0) and (c,0) where {{{c=sqrt(a^2+b^2)}}}

    So we calculate {{{c=sqrt(a^2+b^2)=sqrt(1^2+1^2)=sqrt(1+1)=sqrt(2)}}}

    and the foci are ({{{-sqrt(2)}}},0) and ({{{sqrt(2)}}},0) 


5.  The defining rectangle has the 4 corners 
    (-1,1), (-1,-1), (1,-1), and (1,1)

6.  The asymptotes are the extended diagonals of the defining rectangle
    and have equations {{{y}}}{{{""=""}}}{{{"" +- (1/1)}}}{{{x}}}, or
    {{{y}}}{{{""=""}}}{{{"" +- x}}}

So we draw the defining rectangle and its extended diagonals:

{{{drawing(400,400,-4,4,-4,4, graph(400,400,-4,4,-4,4),

green(rectangle(-1,-1,1,1),line(-1,1,1,1),line(-1,-1,1,-1),
line(-5,-5,5,5),line(-5,5,5,-5))

)   )}}}
    
Now we sketch in the hyperbola:

{{{drawing(400,400,-4,4,-4,4, graph(400,400,-4,4,-4,4,sqrt(x^2-1)),
graph(400,400,-4,4,-4,4,-sqrt(x^2-1)),
green(rectangle(-1,-1,1,1),line(-1,1,1,1),line(-1,-1,1,-1),
line(-5,-5,5,5),line(-5,5,5,-5))

)   )}}}

Edwin</pre>