How do you find all the critical points to graph #x^2 - 9y^2 + 2x - 54y + 80 = 0# including vertices, foci and asympotes?

1 Answer
Oct 19, 2016

Answer:

Please see the explanation.

Explanation:

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#