A hyperbola has vertices at (0, ±3) and foci at (0, ±7). Find the equations of the asymptotes.
3 answers
N/A
since the foci lie on the y-axis, we have
y^2/b^2 - x^2/a^2 = 1
a=3
c=7
so b^2 = 40
The asymptotes are
y = ±b/a x = ±√40/3 x
y^2/b^2 - x^2/a^2 = 1
a=3
c=7
so b^2 = 40
The asymptotes are
y = ±b/a x = ±√40/3 x
equation of hyperbola is
x^2 / 40 - y^2 / 9 = -1
equations of asymptotes:
9x^2 - 40y^2 = 0
(3x - √40y)(3x + √40y) = 0
y = ± 3/√40 x
x^2 / 40 - y^2 / 9 = -1
equations of asymptotes:
9x^2 - 40y^2 = 0
(3x - √40y)(3x + √40y) = 0
y = ± 3/√40 x