How do you find all the critical points to graph #9x^2-16y^2+18x+160y-247=0# including vertices, foci and asymptotes?

1 Answer
Oct 21, 2016

Answer:

The center #(-1, 5)#
Vertices are #(-1, 2)# and #(-1,8)#
Foci: #(-1, 0)# and #(-1,10)#
Asymptotes:#y = 3/4(x - -1) + 5 and y = -3/4(x - -1) + 5 #

Explanation:

Hyperbolas standard forms:
#(x - h)^2/a^2 - (y -k)^2/b^2 = 1# [1]
#(y - k)^2/a^2 - (x -h)^2/b^2 = 1#[2]

Add #247 + 9h^2 - 16k^2# to both sides:

#9x^2 + 18x + 9h^2 - 16y^2 + 160y - 16k^2 = 247 + 9h^2 - 16k^2#

Factor 9 and 16 from two quadratics on the left:

#9(x^2 + 2x + h^2) - 16(y^2 - 10y - k^2) = 247 + 9h^2 - 16k^2#

Complete the squares, #h = -1, h^2 = 1, k = 5, and k^2 = 25#

#9(x - -1)^2 - 16(y - 5)^2 = 247 + 9(1) - 16(25)#

#9(x - -1)^2 - 16(y - 5)^2 = -144#

Divide both sides by -144:

#(y - 5)^2/3^2 - (x - -1)^2/4^2 = 1#

This is type [2]

The center, #(h, k)# is at #(-1, 5)#

To find the vertices, let #x = -1#

#(y - 5)^2/3^2 - (-1 - -1)^2/4^2 = 1#

#(y - 5)^2/3^2 = 1#

#(y - 5)^2 = 3^2 #

#y - 5 = +-3^2 #

#y = 5 + 3 = 8# and #y = 5 - 3 = 2#

Vertices are #(-1, 2)# and #(-1,8)#

The focal length, c,:

#c^2 = a^2 + b^2 = 3^2 + 4^2 = 25#

#c = 5

#y = k - c = 5 - 5 = 0# and #y = k + c = 5 + 5 = 10

Foci: #(-1, 0)# and #(-1,10)#

Asymptotes:

#y - y_0 = +-a/b(x - x_0)#

#y = 3/4(x - -1) + 5 and y = -3/4(x - -1) + 5 #