Add h^2 - 9k^2 - 80h2−9k2−80 to both sides:
x^2 + 2x + h^2 - 9y^2 - 54y - 9k^2 = h^2 - 9k^2 - 80x2+2x+h2−9y2−54y−9k2=h2−9k2−80
Set the right side of the pattern (x - h)^2 = x^2 - 2hx + h^2(x−h)2=x2−2hx+h2 equal to the terms x^2 + 2x + h^2x2+2x+h2:
x^2 - 2hx + h^2 = x^2 + 2x + h^2x2−2hx+h2=x2+2x+h2
Solve for hh and h^2h2:
-2hx + h^2 = 2x + h^2−2hx+h2=2x+h2
-2hx = 2x−2hx=2x
h = -1h=−1 and h^2 = 1h2=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^2h2 on the right:
(x - -1)^2 - 9y^2 - 54y - 9k^2 = 1 - 9k^2 - 80(x−−1)2−9y2−54y−9k2=1−9k2−80
Factor -9 from the y terms:
(x - 1)^2 - 9(y^2 + 6y + k^2) = 1 - 9k^2 - 80(x−1)2−9(y2+6y+k2)=1−9k2−80
Set the right side of the pattern (y - k)^2 = y^2 - 2ky + k^2(y−k)2=y2−2ky+k2 equal to the terms y^2 + 6y + k^2y2+6y+k2:
y^2 - 2ky + k^2= y^2 + 6y + k^2y2−2ky+k2=y2+6y+k2
Solve for kk and k^2k2:
-2ky= 6y−2ky=6y
k = -3k=−3 and k^2 = 9k2=9
Substitute the left side on the pattern (along with the value of k) into the ()s on the left and 9 for k^2k2 on the right:
(x - -1)^2 - 9((y - -3)^2) = 1 - 9(9) - 80(x−−1)2−9((y−−3)2)=1−9(9)−80
Simplify the right side:
(x - -1)^2 - 9((y - -3)^2) = -160(x−−1)2−9((y−−3)2)=−160
Divide both sides by -160:
9((y - -3)^2)/160 - (x - -1)^2/160 = 19(y−−3)2160−(x−−1)2160=1
Write denominators as squares:
(y - -3)^2/((4sqrt(10))/3)^2 - (x - -1)^2/(4sqrt10)^2 = 1(y−−3)2(4√103)2−(x−−1)2(4√10)2=1
Center:(-1,-3)(−1,−3)
To find the vertices, make the negative term disappear by setting x = -1x=−1:
(y - -3)^2/((4sqrt(10))/3)^2 = 1(y−−3)2(4√103)2=1
(y - -3)^2 = ((4sqrt(10))/3)^2(y−−3)2=(4√103)2
y - -3 = (4sqrt(10))/3y−−3=4√103 and y - -3 = -(4sqrt(10))/3y−−3=−4√103
y = -3 + (4sqrt(10))/3y=−3+4√103 and y = -3 -(4sqrt(10))/3y=−3−4√103
Vertices:(-1, -3 + (4sqrt(10))/3)(−1,−3+4√103) and (-1, -3 -(4sqrt(10))/3)(−1,−3−4√103)
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=√1609+160
c = sqrt(160/9 + 160)c=√1609+160
c = 40/3c=403
The foci are at : (-1, -3 + 40/3)(−1,−3+403) and (-1, -3 - 40/3)(−1,−3−403)
Foci:(-1, 31/3)(−1,313) and (-1, - 49/3)(−1,−493)
To find the asymptotes, please observe that dividing the first denominator by the second gives the slopes of 1/313 and -1/3−13. All that is left to do is to force the lines with those two slopes, through the center point.
y = 1/3(x - -1) - 3y=13(x−−1)−3
y = -1/3(x - -1) - 3y=−13(x−−1)−3
