Add h^2 - 9k^2 - 80 to both sides:
x^2 + 2x + h^2 - 9y^2 - 54y - 9k^2 = h^2 - 9k^2 - 80
Set the right side of the pattern (x - h)^2 = x^2 - 2hx + h^2 equal to the terms x^2 + 2x + h^2:
x^2 - 2hx + h^2 = x^2 + 2x + h^2
Solve for h and h^2:
-2hx + h^2 = 2x + h^2
-2hx = 2x
h = -1 and h^2 = 1
Substitute the left side of the pattern (along with the value of h) for the 3 terms on the left and substitute 1 for h^2 on the right:
(x - -1)^2 - 9y^2 - 54y - 9k^2 = 1 - 9k^2 - 80
Factor -9 from the y terms:
(x - 1)^2 - 9(y^2 + 6y + k^2) = 1 - 9k^2 - 80
Set the right side of the pattern (y - k)^2 = y^2 - 2ky + k^2 equal to the terms y^2 + 6y + k^2:
y^2 - 2ky + k^2= y^2 + 6y + k^2
Solve for k and k^2:
-2ky= 6y
k = -3 and k^2 = 9
Substitute the left side on the pattern (along with the value of k) into the ()s on the left and 9 for k^2 on the right:
(x - -1)^2 - 9((y - -3)^2) = 1 - 9(9) - 80
Simplify the right side:
(x - -1)^2 - 9((y - -3)^2) = -160
Divide both sides by -160:
9((y - -3)^2)/160 - (x - -1)^2/160 = 1
Write denominators as squares:
(y - -3)^2/((4sqrt(10))/3)^2 - (x - -1)^2/(4sqrt10)^2 = 1
Center:(-1,-3)
To find the vertices, make the negative term disappear by setting x = -1:
(y - -3)^2/((4sqrt(10))/3)^2 = 1
(y - -3)^2 = ((4sqrt(10))/3)^2
y - -3 = (4sqrt(10))/3 and y - -3 = -(4sqrt(10))/3
y = -3 + (4sqrt(10))/3 and y = -3 -(4sqrt(10))/3
Vertices:(-1, -3 + (4sqrt(10))/3) and (-1, -3 -(4sqrt(10))/3)
To find the focal distance, c, take the square root of the sum of the squares of the denominators:
c = sqrt(160/9 + 160)
c = sqrt(160/9 + 160)
c = 40/3
The foci are at : (-1, -3 + 40/3) and (-1, -3 - 40/3)
Foci:(-1, 31/3) and (-1, - 49/3)
To find the asymptotes, please observe that dividing the first denominator by the second gives the slopes of 1/3 and -1/3. All that is left to do is to force the lines with those two slopes, through the center point.
y = 1/3(x - -1) - 3
y = -1/3(x - -1) - 3
