Question 531225
A hyperbola with its Center (-3,2), transverse axis parallel to the y-axis, passing through (1,7), the asymptotes are perpendicular to each other.
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Standard form of equation for a hyperbola with vertical transverse axis, as in this case:
(y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
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Took me awhile to figure out that if the asymptotes are perpendicular to each other their slopes must be equal to ±1.
slope=a/b=±1
so, a=b or a=-b
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Given equation:
(y-2)^2/a^2-(x+3)^2/b^2=1
can be rewritten:
(y-2)^2/a^2-(x+3)^2/a^2=1
using given point on parabola (1,7) to solve for a
(7-2)^2/a^2-(1+3)^2/a^2=1
25/a^2-16/a^2=1
9/a^2=1
a^2=9
a=3
b=a=3
Equation of given hyperbola:
(y-2)^2/9-(x+3)^2/9=1
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Asymptotes:
Asymptotes are straight lines thru the center of the hyperbola of the standard form:
y=mx+b, m=slope, b=y-intercept
for slope=1
2=1*-3+b
b=5
equation of asymptote
y=x+5
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for slope=-1
2=-1*-3+b
b=-1
equation of asymptote
y=-x-1
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See graph as a visual check on answers:
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y=±(9+(x+3)^2)^.5+2
{{{ graph( 300, 300, -10, 10, -10, 10,(9+(x+3)^2)^.5+2,-(9+(x+3)^2)^.5+2,x+5,-x-1) }}}
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