Asked by Stefan
Find the center, vertices, foci, lengths of transverse and conjugate axis and the asymptotes of the hyperbola with equation:
9x² - 16y² - 36x - 96y - 252 = 0
9x² - 16y² - 36x - 96y - 252 = 0
Answers
Answered by
oobleck
9x^2 - 36x - 16y^2 - 96y = 252
9(x^ - 2)^2 - 16(y+3)^2 = 144
(x-2)^2/16 - (y+3)^2/9 = 1
so we have, knowing that the axis of symmetry is the line y = -3
center: (2,-3)
vertices (a=4): (2±4,-3)
foci (c=5): (2±5,-3)
ftransverse axis = 2a = 8
conjugate axis = 2b = 6
asymptotes (±b/a) : y+3 = ±3/4 (x-2)
9(x^ - 2)^2 - 16(y+3)^2 = 144
(x-2)^2/16 - (y+3)^2/9 = 1
so we have, knowing that the axis of symmetry is the line y = -3
center: (2,-3)
vertices (a=4): (2±4,-3)
foci (c=5): (2±5,-3)
ftransverse axis = 2a = 8
conjugate axis = 2b = 6
asymptotes (±b/a) : y+3 = ±3/4 (x-2)
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