Question 281903
what are the slopes of the asymptotes of the hyperbola with equation 
{{{4x^2-y^2-8x+2y-1=0}}}
<pre><font size = 4 color = "indigo"><b>
First get it in standard form, which is either

{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}} if the hyperbola opens right and left, 

or

{{{(y-k)^2/a^2-(x-h)^2/b^2=1}}} if the hyperbola opens upward and downward.

{{{4x^2-y^2-8x+2y-1=0}}}

Get the loose number -1 off the left side

{{{4x^2-y^2-8x+2y=1}}}

Get the {{{x}}} term next to the {{{x^2}}} term.
Get the {{{y}}} term next to the {{{y^2}}} term.

{{{4x^2-8x-y^2+2y=1}}}

Factor the coefficient of {{{x^2}}} out of the 
first two terms on the left. 
Farctor the coefficient of {{{y^2}}} out of the 
last two terms on the left. 

{{{4(x^2-2x)-1(y^2-2y)=1}}}

Complete the square on {{{(x^2-2x)}}} by multiplying
the coefficient of x, which is -2 by {{{1/2}}} getting -1,
and then squaring -1, getting +1.  And we add that inside the
first parentheses.  However since there is a 4 in fromt of the
first parentheses, adding +1 inside the parentheses amounts
to adding +4*1 or +4 to the left side, so we must add +4 
to the right side:

{{{4(x^2-2x+red(1))-1(y^2-2y)=1+red(4)}}}


Complete the square on {{{(y^2-2y)}}} by multiplying
the coefficient of y, which is -2 by {{{1/2}}} getting -1,
and then squaring -1, getting +1.  And we add that inside the
second parentheses.  However since there is a -1 in fromt of the
second parentheses, adding +1 inside the parentheses amounts
to adding {{{1*-1}}} or -1 to the left side, so we must add -1 
to the right side:

{{{4(x^2-2x+red(1))-1(y^2-2y+green(1))=1+red(4)+green((-1))}}}

Factor the parentheses as squares of binomials, and combine
the numbers on the right:

{{{4(x-1)^2-(y-1)^2=4}}}

Get a 1 on the right by dividing through by 4

{{{4(x-1)^2/4-(y-1)^2/4=4/4}}}

{{{(x-1)^2/1-(y-1)^2/4=1}}}

Since the variable x comes first in the standard form, the
hyperbola opens right and left.

So we compare that to:

{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}

The center is at (1,1).  So we plot the center:

{{{drawing(400,400,-6,8,-6,8, graph(400,400,-6,8,-6,8),

line(1+.1,1,1-.1,1), line(1,1+.1,1,1-.1), line(1+.1,1+.1,1-.1,1-.1), line(1+.1,1-.1,1-.1,1+.1)  )}}}

{{{a^2=1}}} so {{{a=1}}}, the semi-transverse axis is 1 unit
long, so we draw the complete transverse axis right and left
1 unit from the center, that is, the tranverse axis is the 
horizontal green line below:

{{{drawing(400,400,-6,8,-6,8, graph(400,400,-6,8,-6,8),
green(line(0,1,2,1)),
line(1+.1,1,1-.1,1), line(1,1+.1,1,1-.1), line(1+.1,1+.1,1-.1,1-.1), line(1+.1,1-.1,1-.1,1+.1)  )}}} 

The two vertices are the endpoints of the transverse axis, so the
vertices are the two points 

(0,1) and (2,1)


{{{b^2=4}}} so {{{b=2}}}, the semi-conjugate axis is 2 units
long, so we draw the complete conjugate axis up and down
2 units from the center, that is, the conjugate axis is the 
vertical green line below:

{{{drawing(400,400,-6,8,-6,8, graph(400,400,-6,8,-6,8),
green(line(0,1,2,1)), green(line(1,-1,1,3)),
line(1+.1,1,1-.1,1), line(1,1+.1,1,1-.1), line(1+.1,1+.1,1-.1,1-.1), line(1+.1,1-.1,1-.1,1+.1)  )}}} 

Now we draw in the defining rectangle

{{{drawing(400,400,-6,8,-6,8, graph(400,400,-6,8,-6,8),
green(line(0,1,2,1)), green(line(1,-1,1,3)),
line(1+.1,1,1-.1,1), line(1,1+.1,1,1-.1), line(1+.1,1+.1,1-.1,1-.1), line(1+.1,1-.1,1-.1,1+.1),
rectangle(0,-1,2,3)  )}}} 

Now we can draw the asymptotes which are the extended diagonals
of the defining rectangle:

{{{drawing(400,400,-6,8,-6,8, graph(400,400,-6,8,-6,8,2x-1),
green(line(0,1,2,1)), green(line(1,-1,1,3)), graph(400,400,-6,8,-6,8,-2x+3),
line(1+.1,1,1-.1,1), line(1,1+.1,1,1-.1), line(1+.1,1+.1,1-.1,1-.1), line(1+.1,1-.1,1-.1,1+.1),
rectangle(0,-1,2,3)  )}}}

and we can sketch in the hyperbola:

{{{drawing(400,400,-6,8,-6,8, graph(400,400,-6,8,-6,8,2x-1,1+2sqrt((x-1)^2-1)),
graph(400,400,-6,8,-6,8,-2x+3,1-2sqrt((x-1)^2-1)),

green(line(0,1,2,1)), green(line(1,-1,1,3)),
line(1+.1,1,1-.1,1), line(1,1+.1,1,1-.1), line(1+.1,1+.1,1-.1,1-.1), line(1+.1,1-.1,1-.1,1+.1),
rectangle(0,-1,2,3)

  )}}} 

The slopes of the asymptotes are {{{""+-b/a = ""+- 2/1=""+- 2}}}

They go through the point which is the center of the hyperbola (1,1)

{{{y-y[1]=m(x-x[1])}}}

Using m=2

{{{y-1=2(x-1)}}}

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

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

Using m=-2

{{{y-1=-2(x-1)}}}

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

{{{y=-2x+3}}}

To find the foci. we find c using

{{{c^2=a^2+b^2}}}
{{{c^2=1^2+2^2}}}
{{{c^2=1+4}}}
{{{c^2=5}}}
{{{c=""+-sqrt(5)}}}

The foci are c units right and left of the center:

They are (1-{{{sqrt(5)}}},1) and (1+{{{sqrt(5)}}},1)
-sqrt(5)
I'll draw the foci in:

{{{drawing(400,400,-6,8,-6,8, graph(400,400,-6,8,-6,8,2x-1,1+2sqrt((x-1)^2-1)),
graph(400,400,-6,8,-6,8,-2x+3,1-2sqrt((x-1)^2-1)),

green(line(0,1,2,1)), green(line(1,-1,1,3)),
line(1-sqrt(5)+.1,1,1-sqrt(5)-.1,1), line(1-sqrt(5),1+.1,1-sqrt(5),1-.1), line(1-sqrt(5)+.1,1+.1,1-sqrt(5)-.1,1-.1), line(1-sqrt(5)+.1,1-.1,1-sqrt(5)-.1,1+.1),
rectangle(0,-1,2,3),

line(1+.1,1,1-.1,1), line(1,1+.1,1,1-.1), line(1+.1,1+.1,1-.1,1-.1), line(1+.1,1-.1,1-.1,1+.1),

line(1+sqrt(5)+.1,1,1+sqrt(5)-.1,1), line(1+sqrt(5),1+.1,1+sqrt(5),1-.1), line(1+sqrt(5)+.1,1+.1,1+sqrt(5)-.1,1-.1), line(1+sqrt(5)+.1,1-.1,1+sqrt(5)-.1,1+.1)





  )}}}  





Edwin</pre>