SOLUTION: Find the equation of the hyperbola with vertices at (-1, -1) and (3, -1) and an equation of asymptote (𝑥−1)/2=(𝑦+1)/3

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the equation of the hyperbola with vertices at (-1, -1) and (3, -1) and an equation of asymptote (𝑥−1)/2=(𝑦+1)/3      Log On


   

Question 1194765: Find the equation of the hyperbola with vertices at (-1, -1) and (3, -1) and an equation of asymptote (𝑥−1)/2=(𝑦+1)/3
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

the equation of the hyperbola :
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1 where (h,k) is the center, a and b are the lengths of the semi-major and the semi-minor axes
given:
vertices at (-1, -1) and (3,+-1)
The distance between the vertices is 2a. Since y=-1, distance between x=-1 and x=3 is 4 units
=> so, 2a=4=> a=2
since center is midpoint between vertices, C(%28-1%2B3%29%2F2,%28-1-1%29%2F2)=(1,1)
=>h=1 and k=-1

equation of asymptote %28x-1%29%2F2+=+%28y%2B1%29%2F3=>y+=+%283%2F2%29x+-+5%2F2=> b%2Fa=3%2F2
since a=2=>b%2F2=3%2F2=>b=3

the equation of the hyperbola :
%28x-1%29%5E2%2F4-%28y%2B1%29%5E2%2F9=1