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Question 598756: Find the equation of hyperbola whose asymptotes are 5x+3y=-1 and 5x-3y=11, and which passes through (4,-2)
I found the slopes of the equations -5/3 and 5/3. I do not know what to do next. Please help!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the equation of hyperbola whose asymptotes are 5x+3y=-1 and 5x-3y=11, and which passes through (4,-2)
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Asymptotes are straight lines which go thru the center and are of the standard form:y=mx+b, m=slope,
..
Finding coordinates of center with given asymptote equations:
5x+3y=-1
5x-3y=11
subtract
6y=-12
y=-2
..
add equations
10x=10
x=1
..
So, center is (1,-2) (point of intersection of asymptotes.)
Given point (4,-2) shows that hyperbola has a horizontal transverse axis which would have the following standard form of equation:
(x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
slope of asymptote=±5/3=b/a
b=±5a/3
b^2=25a^2/9
using coordinates of center(1,-2) and given point on hyperbola(4,-2):
(4-1)^2/a^2-(-2+2)^2/b^2=1
9/a^2-0/b^2=1
a^2=9
a=3
b^2=25*9/9=25
equation:
(x-1)^2/9-(y+2)^2/25=1
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