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#
