SOLUTION: Find the coordinates of the center, vertices, foci and equation of the asymptotes of the following hyperbolas: a. 36x2 - 25y2 - 72x + 50y - 889 = 0 b. x2 - 4y2 - x + 12

Swap the middle two terms to get like lettered terms together:



Factor 36 out of the two terms in x:  (Factor out 36, not 36x)



Factor -25 out of the two terms in y:  (Factor out -25, not -25y)



Add 889 to both sides:



Complete the square inside the first parentheses:

1. To the side, multiply the coefficient of x, which is -2, by 1/2, getting -1.
2. Square this value, get +1
3. Add then subtract it "+1-1" inside the first parentheses

 

Complete the square inside the second parentheses:

1. To the side, multiply the coefficient of y, which is -2, by 1/2, getting -1.
2. Square this value, get +1
3. Add then subtract it "+1-1" inside the second parentheses



Factor the first three terms inside the first parentheses using a large
parentheses with small parentheses inside:

 

Factor the first three terms inside the second parentheses using a large
parentheses with small parentheses inside:



Write the products of binomials with themselves as the square of one
binomial:



Remove the larger parentheses by distributing, leaving the smaller
parentheses intact:



Combine the two constant terms on the left



Add 11 to both sides



Divide each term on both sides by 900 to make the right side become 1:



Divide the numerators and denominators by the coefficients on top:



Compare this to



and we see that

h=1, k=1, a2=25, a=5, b2=36, b=6.

The center is (h,k) = (1,1).

We plot the center.  We draw the transverse axis which is a=5 units on both
sides of the center.  Then we draw the conjugate axis which is b=6 units vertically above and below the center.




We draw the defining rectangle so that the two axes bisect it:




We draw and extend the two diagonals of the defining rectangle, which are the
asymptotes of the hyperbola:




Now we sketch in the hyperbola:



The center is (1,1)
The vertices are (-4,1) and (6,1)
To find the foci, we use the Pythagorean formula
to find their distance from the center







That is added to and subtracted from the x coordinate of the center to get
the coordinates of the foci. The foci are on the extension of the transverse
axes, shown below, and have the coordinates:

 



Edwin