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| Question 758896:  Determine the equation of the hyperbola whose asymptotes are x±2y=0 and which passes through (4,3)
 Answer by lwsshak3(11628)
      (Show Source): 
You can put this solution on YOUR website! Determine the equation of the hyperbola whose asymptotes are x±2y=0 and which passes through (4,3) ***
 Hyperbola has a vertical transverse axis since it passes thru (4,3)
 Its standard form of equation: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of center
 Asymptotes are two straight line equations that intersect at center, y=mx+b, m=slope, b=y-intercept
 Equation of asymptote with positive slope:
 x-2y=0
 2y=x
 y=x/2
 m=1/2, b=0
 ..
 Equation of asymptote with negative slope:
 x+2y=0
 2y=-x
 y=-x/2
 m=-1/2, b=0
 This means center is at (0,0)
 ..
 slopes of asymptotes with vertical transverse axis
 =±a/b=±1/2
 b=±2a
 b^2=4a^2
 ..
 solving for a^2 and b^2 using coordinates of given point(4,3)
 (y-k)^2/a^2-(x-h)^2/b^2=1
 y^2/a^2-x^2/4a^2=1
 9/a^2-16/4a^2=1
 9/a^2-4/a^2=1
 5/a^2=1
 a^2=5
 b^2=4a^2=20
 Equation:
  
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