Asked by jayson
what is the standard equation given asymptotes are y=1/3(x+5) and y=-1/3(x-7) with foci (1,12)
Answers
Answered by
Steve
The asymptotes intersect at the center: (1,2)
The foci are at x=1, so the axis is vertical, giving
(y-2)^2/a^2 - (x-1)^2/b^2 = 1
If a focus is at (1,12), c=10
c^2 = a^2+b^2
The slope of the asymptotes is a/b = 1/3, so b=3a
c^2 = a^2+b^2 = 100, so
a^2=10, b^2=90
(y-2)^2/10 - (x-1)^2/90 = 1
verify at
http://www.wolframalpha.com/input/?i=hyperbola+(y-2)%5E2%2F10-+(x-1)%5E2%2F90+%3D+1
The foci are at x=1, so the axis is vertical, giving
(y-2)^2/a^2 - (x-1)^2/b^2 = 1
If a focus is at (1,12), c=10
c^2 = a^2+b^2
The slope of the asymptotes is a/b = 1/3, so b=3a
c^2 = a^2+b^2 = 100, so
a^2=10, b^2=90
(y-2)^2/10 - (x-1)^2/90 = 1
verify at
http://www.wolframalpha.com/input/?i=hyperbola+(y-2)%5E2%2F10-+(x-1)%5E2%2F90+%3D+1
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