SOLUTION: How do I put this equation in standard hyperbola form AND then find the coordinates of the foci? My biggest problem is putting it in standard hyperbola form because I am bad at fac

-16x² + 25y² - 32x - 250y + 209 = 0

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:

      -16x² + 25y² - 32x - 250y + 209 = 0

We get the 209 on the other side as a -209:

            -16x² + 25y² - 32x - 250y = -209

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

            -16x² - 32x + 25y² - 250y = -209

Write it this way: 

        [-16x² - 32x] + [25y² - 250y] = -209
 
The coefficient of x² is -16, so let's factor that out
in the 1st bracket.  (Remember to change the sign when
factoring out a negative.  That's why we have +2x 
and not -2x]:
 
        [-16(x² + 2x)] + [25y² - 250y] = -209

The coefficient of y² is 15, so let's factor that out
in the 2nd bracket.  

       [-16(x² + 2x)] + [25(y² - 10y)] = -209

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

           -16(x² + 2x) + 25(y² - 10y) = -209

Next we want to make those two binomials into trinomials.

We skip some space after those binomials 

   -16(x² + 2x    ) + 25(y² - 10y    ) = -209

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 2 so we take half of it, getting 1,
then we square 1, getting 1² or 1, but wait!  See the -16 in 
front of the first parentheses?  If we put a 1 in that box,
It will get multiplied by the -16 in front of the parentheses.
In other words putting a 1 in that first space will in effect 
amount to the same as adding -16 times 1 or -16 to the left side,
not just 1.  So we have to add -16(1) to the right side to offset
adding 1 inside that parentheses on the left since it will be 
multiplied by the -16, so we so we add 1 in the first space, but 
we have to add -16 to the other side of the equation:
 
  -16(x² + 2x + 1) + [25(y² - 10y    ) = -209 - 16(1)

Now let's figure out what number goes in the second space.
 
The coefficient of y is -10 so we take half of it, getting -5,
then we square -5, getting (-5)² or 25, but wait!  See the 25 in 
front of the second parentheses?  If we put a 25 in that box,
It will get multiplied by the 25 in front of the parentheses.
In other words putting a 25 in that second box will in effect 
amount to the same as adding 25 times 25 or 625 to the left side,
not just 25.  So we have to add 25(25) to the right side to offset
adding 25 inside that parentheses since it will be multiplied
by the 25, so we have:

  -16(x² + 2x + 1) + [25(y² - 10y + 25) = -209 - 16(1) + 25(25)

Notice that what's in the first parentheses,
x²+2x+1 factors as (x-1)(x-1) or (x-1)²

Also notice that what's in the second parentheses
y²-10y+25 factors as (y-5)(y-5) or (y+1)².

So this

   -16(x² + 2x + 1) + 25(y² - 10y + 25) = -209 - 16(1) + 25(25)

becomes this
   
                   -16(x+1)² + 25(y-5)² = 400

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 400:

-16(x+1)²   25(y-5)²    400
————————— + ———————— = ————— 
   400        400       400

And that simplifies to:

    (x+1)²   (y-5)²    
 - —————— + ——————— = 1
     25       16       
 
Let's write the positive term first:

    (y-5)²   (x+1)²    
   —————— - ——————— = 1
     16       25

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=-1, k=5, a²=16, b²=25.

The center is (h,k) = (-1,5)

Plot it:



Since a² = 16, a = 4
Since b² = 25, b = 5

a = 4 is the semi-transverse axis's length, so draw the vertical
transverse axis 2a or 8 units long with the center as the midpoint.
We also draw the horizontal conjugate axis 2b or 10 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:



The vertices are (-1,1) and (-1,9)

We only need to find the coordinates of the two foci.

The foci are points which are c units from the center and colinear
with the vertices and the center:

We calculate c from:

c² = a² + b²
c² = 4² + 5²
c² = 16 + 25
c² = 41
 c = √41 about 6.4
                        __              __
So the foci are F(-1,5+√41) and F(-1,5-√41).




Edwin