Question 281731
what are the slopes of the asymptotes of the hyperbola with equation 
{{{4x^2-y^2+8x-6y=9}}}
<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, 

and the slopes of the asymptotes are {{{""+-b/a}}}

or

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

and the slopes of the asymptotes are {{{""+-a/b}}}



{{{4x^2-y^2+8x-6y=9}}}

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

{{{4x^2+8x-y^2-6y=9}}}

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+6y)=9}}}

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+6y)=9+red(4)}}}


Complete the square on {{{(y^2+6y)}}} by multiplying
the coefficient of y, which is 6 by {{{1/2}}} getting 3,
and then squaring 3, getting 9.  And we add that inside the
second parentheses.  However since there is a -1 in fromt of the
second parentheses, adding 9 inside the parentheses amounts
to adding {{{9*-1}}} or -9 to the left side, so we must add -9 
to the right side:

{{{4(x^2+2x+red(1))-1(y^2+6y+green(9))=9+red(4)+green((-9))}}}

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

{{{4(x+1)^2-(y+3)^2=4}}}

Get a 1 on the right by dividing through by 4

{{{4(x+1)^2/4-(y+3)^2/4=4/4}}}

{{{(x+1)^2/1-(y+3)^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,-3).  So we plot the center:

{{{drawing(400,400,-8,6,-10,4, graph(400,400,-8,6,-10,4),

line(-1+.1,-3,-1-.1,-3), line(-1,-3+.1,-1,-3-.1), line(-1+.1,-3+.1,-1-.1,-3-.1), line(-1+.1,-3-.1,-1-.1,-3+.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,-8,6,-10,4, graph(400,400,-8,6,-10,4),
green(line(-2,-3,0,-3)),
line(-1+.1,-3,-1-.1,-3), line(-1,-3+.1,-1,-3-.1), line(-1+.1,-3+.1,-1-.1,-3-.1), line(-1+.1,-3-.1,-1-.1,-3+.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,-8,6,-10,4, graph(400,400,-8,6,-10,4),
green(line(-2,-3,0,-3)), green(line(-1,-1,-1,-5)),
line(-1+.1,-3,-1-.1,-3), line(-1,-3+.1,-1,-3-.1), line(-1+.1,-3+.1,-1-.1,-3-.1), line(-1+.1,-3-.1,-1-.1,-3+.1)  )}}} 

Now we draw in the defining rectangle

{{{drawing(400,400,-8,6,-10,4, graph(400,400,-8,6,-10,4),
green(line(-2,-3,0,-3)), green(line(-1,-1,-1,-5)),
line(-1+.1,-3,-1-.1,-3), line(-1,-3+.1,-1,-3-.1), line(-1+.1,-3+.1,-1-.1,-3-.1), line(-1+.1,-3-.1,-1-.1,-3+.1),
rectangle(-2,-5,0,-1)


  )}}} 

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

{{{drawing(400,400,-8,6,-10,4, graph(400,400,-8,6,-10,4,2x-1),
graph(400,400,-8,6,-10,4,-2x-5),

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


  )}}} 

and we can sketch in the hyperbola:

{{{drawing(400,400,-8,6,-10,4, graph(400,400,-8,6,-10,4,2x-1,-3+2sqrt((x+1)^2-1)),
graph(400,400,-8,6,-10,4,-2x-5,-3-2sqrt((x+1)^2-1)),

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

  )}}} 

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

Edwin</pre>