25x^2-16y^2-100x+96y+276=

The object is to make it look like this:

 (x-h)²  (y-k)²    
—————— - ——————— = 1
   a²      b²

which is a hyperbola that looks like this )( or:

 (y-k)²  (x-h)²    
—————— + ——————— = 1
   a²      b²

Which has one branch opening upward and the other downward

We start with this:


       25x² - 16y² - 100x + 96y + 276 = 0

We get the 276 on the other side as a -276:

             25x² - 16y² - 100x + 96y = -276

We have to switch the two middle terms on the
left so that the terms are in the order "x², x, y², y".

             25x² - 100x - 16y² + 96y = -276

Write it this way: 

        [25x² - 100x] - [16y² - 96y] = -276
 
The coefficient of x² is 25, so let's factor that out
in the 1st bracket.  
 
        [25(x² - 4x)] - [16y² - 96y] = -276

The coefficient of y² is 16, so let's factor that out
in the 2nd bracket.  

       [25(x² - 4x)] - [16(y² - 6y)] = -276

Now we'll dispense with the brackets and just have parentheses:

           25(x² - 4x) - 16(y² - 6y) = -276

Next we want to make those two binomials into trinomials.

We skip some space after those binomials 

   25(x² - 4x    ) - 16(y² - 6y    ) = -276

so we can add a number in those two spaces to make those 
binomials into trinomials so they'll factor into squares 
of binomials.

Now let's figure out what number goes in the first space.

The coefficient of x is -4 so we take half of it, getting -2,
then we square -2, getting (-2)² or 4, but wait!  See the 25 in 
front of the first parentheses?  If we put a 4 in that space,
It will get multiplied by the 25 in front of the parentheses.
In other words putting a 4 in that first space will in effect 
amount to the same as adding 25 times 4 or 100 to the left side,
not just 4.  So we have to add 25(4) to the right side to offset
adding 4 inside that parentheses on the left since it will be 
multiplied by the 25, so we add 4 in the first space, but 
we have to add 100 to the other side of the equation:
 
   25(x² - 4x + 4) - 16(y² - 6y    ) = -276 + 100

Now let's figure out what number goes in the second space.
 
The coefficient of y is -6 so we take half of it, getting -3,
then we square -3, getting (-3)² or 9, but wait!  See the -16 in 
front of the second parentheses?  If we put a 9 in that space,
It will get multiplied by the -16 in front of the parentheses.
In other words putting a 9 in that second box will in effect 
amount to the same as adding -16 times 9 or -144 to the left side,
not just 9.  So we have to add -144 to the right side to offset
adding 9 inside that parentheses since it will be multiplied
by the -16, so we have:

   25(x² - 4x + 4) - 16(y² - 6y + 9) = -276 + 100 - 144

Notice that what's in the first parentheses,

x²-4x+4 factors as (x-2)(x-2) or (x-2)²

Also notice that what's in the second parentheses

y²-6y+9 factors as (y-3)(y-3) or (y-3)².

So this

   25(x² - 4x + 4) - 16(y² - 6y + 9) = -276 + 100 - 144

becomes this
   
                   25(x-2)² - 16(y-3)² = -320

after substituting their factorization for the parentheses
and combining the terms on the right.

Next we get a 1 on the right by dividing all three terms by -320:

 25(x-2)²   16(y-3)²    -320
————————— - ———————— = —————— 
  -320       -320       -320

Divide top and bottom of the first term by 25
Divide top and bottom of the first term by 16

 (x-2)²     (y-3)²    
———————— - ———————— = 1
-320/25      -20       

Simplifying the signs:

   (x-2)²     (y-3)²    
- ———————— + ———————— = 1
   320/25      20  

Reducing the fraction on the bottom of the first term:

   (x-2)²     (y-3)²    
- ———————— + ———————— = 1
    64/5       20

Let's write the positive term first:

    (y-3)²   (x-2)²    
   —————— - ——————— = 1
     20      64/5

which is in the form:

    (y-k)²  (x-h)²    
   —————— + ——————— = 1
      a²      b²

So the hyperbola has one branch opening upward
and the other downward.

We now have h=2, k=3, a²=20, b²=64/5.

The center is (h,k) = (2,3)

Plot it:


                    __     _ 
Since a² = 20, a = √20 = 2√5
                        _     _
Since b² = 64/5, b = 8/√5 = 8√5/5
      _
a = 2√5 is the semi-transverse axis's length, so draw the vertical
transverse axis 2a or 4√5 units long with the center as the midpoint.
We also draw the horizontal conjugate axis 2b or 16√5/5 units long 
with the center as the midpoint:



Now we draw the defining rectangle with the ends of the transverse
and conjugate axes as midpoints of the sides:



Now draw and extend the diagonals of the defining rectangle
which are the asymptotes of the hyperbola:



Now we can sketch in the hyperbola:


Edwin