Question 1121384
<br>
(1) Move the constant to the other side of the equation; factor out the leading coefficients in x and y:<br>
{{{16(x^2+8x)-9(y^2-20y) = 788}}}<br>
(2) Complete the square in both x and y, remembering to add the same amounts to both sides of the equation:<br>
{{{16(x^2+8x+16)-9(y^2-20y+100) = 788+256-900 = 144}}}<br>
(3) Write the trinomials as binomials squared; divide through by 144 to get "1" on the right side:<br>
{{{(x+4)^2/9-(y-10)^2/16 = 1}}}<br>
or<br>
{{{(x+4)^2/3^2-(y-10)^2/4^2 = 1}}}<br>
This is standard form,
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}.<br>
The center is (h,k) = (-4,10); a=3 is the distance from the center to each vertex; b=4 is the distance from the center to each co-vertex.<br>
The distance from the center to each focus is c, where for a hyperbola {{{c^2 = a^2+b^2}}}.  So c=5.<br>
Answers:
(a) The equation is {{{(x+4)^2/9-(y-10)^2/16 = 1}}}.
(b) The branches of the hyperbola open left and right; The foci are a distance c=5 right or left of the center, at (-9,10) and (1,10).
(c) The slopes of the asymptotes are b/a and -b/a; in this example, 4/3 and -4/3.  Use the point-slope form of the equation of a line using each of those slopes, using the center (-4,10) as the point.