SOLUTION: what are the slopes of the asymptotes of the hyperbola with equation {{{4x^2-y^2+8x-6y=9}}}

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Question 281731: what are the slopes of the asymptotes of the hyperbola with equation
4x%5E2-y%5E2%2B8x-6y=9

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
what are the slopes of the asymptotes of the hyperbola with equation
4x%5E2-y%5E2%2B8x-6y=9

First get it in standard form, which is either

%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1 if the hyperbola opens right and left, 

and the slopes of the asymptotes are %22%22%2B-b%2Fa

or

%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1 if the hyperbola opens upward and downward.

and the slopes of the asymptotes are %22%22%2B-a%2Fb



4x%5E2-y%5E2%2B8x-6y=9

Get the x term next to the x%5E2 term.
Get the y term next to the y%5E2 term.

4x%5E2%2B8x-y%5E2-6y=9

Factor the coefficient of x%5E2 out of the 
first two terms on the left. 
Farctor the coefficient of y%5E2 out of the 
last two terms on the left. 

4%28x%5E2%2B2x%29-1%28y%5E2%2B6y%29=9

Complete the square on %28x%5E2%2B2x%29 by multiplying
the coefficient of x, which is 2 by 1%2F2 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%28x%5E2%2B2x%2Bred%281%29%29-1%28y%5E2%2B6y%29=9%2Bred%284%29


Complete the square on %28y%5E2%2B6y%29 by multiplying
the coefficient of y, which is 6 by 1%2F2 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%2A-1 or -9 to the left side, so we must add -9 
to the right side:



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

4%28x%2B1%29%5E2-%28y%2B3%29%5E2=4

Get a 1 on the right by dividing through by 4

4%28x%2B1%29%5E2%2F4-%28y%2B3%29%5E2%2F4=4%2F4

%28x%2B1%29%5E2%2F1-%28y%2B3%29%5E2%2F4=1

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

So we compare that to:

%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1

The center is at (-1,-3).  So we plot the center:



a%5E2=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:

 

b%5E2=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:

 

Now we draw in the defining rectangle

 

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

 

and we can sketch in the hyperbola:

 

The slopes of the asymptotes are %22%22%2B-b%2Fa+=+%22%22%2B-+2%2F1=%22%22%2B-+2

Edwin