SOLUTION: Convert the equation to standard form. Locate the foci and find the equation of the asymptotes. 16x2 + 128x - 9y2 + 180y - 788 = 0

Algebra ->  Finance -> SOLUTION: Convert the equation to standard form. Locate the foci and find the equation of the asymptotes. 16x2 + 128x - 9y2 + 180y - 788 = 0       Log On


   

Question 1121384: Convert the equation to standard form. Locate the foci and find the equation of the asymptotes.
16x2 + 128x - 9y2 + 180y - 788 = 0
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Use ^ for exponents.

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


(1) Move the constant to the other side of the equation; factor out the leading coefficients in x and y:

16%28x%5E2%2B8x%29-9%28y%5E2-20y%29+=+788

(2) Complete the square in both x and y, remembering to add the same amounts to both sides of the equation:

16%28x%5E2%2B8x%2B16%29-9%28y%5E2-20y%2B100%29+=+788%2B256-900+=+144

(3) Write the trinomials as binomials squared; divide through by 144 to get "1" on the right side:

%28x%2B4%29%5E2%2F9-%28y-10%29%5E2%2F16+=+1

or

%28x%2B4%29%5E2%2F3%5E2-%28y-10%29%5E2%2F4%5E2+=+1

This is standard form,
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1.

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.

The distance from the center to each focus is c, where for a hyperbola c%5E2+=+a%5E2%2Bb%5E2. So c=5.

Answers:
(a) The equation is %28x%2B4%29%5E2%2F9-%28y-10%29%5E2%2F16+=+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.