SOLUTION: I cannot seem to find the correct standard equation, foci, and asymptotes of the following hyperbola: {{{ x^2-4y^2-6x-16y-11=0 }}} I need to write it in this kind of form: {

Rearrange so that the two terms in x are next to
each other and the two terms in y are next to
each other, and 11 to both sides:
sides:



The coefficient of  is already 1, so
we don't factor anything out of it, just enclose the
first two terms in parentheses:



Now factor only -4 (not y) out of the last two terms
on the left.



Complete the square in each parentheses.  

In the first parentheses multiply the  coefficient of
x by , getting .  Then we square ,
getting .  So we add 9 to both sides:





Now we factor the first parentheses as 
or 



In the second parentheses multiply the  coefficient of
y by , getting .  Then we square ,
getting .  So we add 4 inside the second
parentheses.

But since there is a  factor in front of the second
parentheses, when we add  inside the second parentheses
we are really adding  times  or  to the
left side, so we must add  to the right side too.





Now we factor the second parentheses as 
or 



Next we get a  on the right side by dividing
through by 





Now to get rid of the 4 on top on the second term, we
divide top and bottom by 4







Can you go from here? 

Compare to



So , , , 

So center = (h,k) = (3,-2)

The two vertices are a=2 units left and right of the
center, so they are (1,-2) and (5,-2)

The two foci are  units left and right of the center.

First we calculate  using







So the two foci are  units left and right of the
center, so they are (,-2) and (,-2)

The asymptotes have slopes which are ± or
±

And the go through the center (3,-2), so we use the
point slope formula:

For the asymptote that has slope :



Multiply both sides by 2







---------------

For the asymptote that has slope :



Multiply both sides by -2






 
Edwin