SOLUTION: Could someone please help me with writing a conic equation in standard form for a hyperbola with vertices at (1,6) and (1,-2) and asymptotes at y=(4/3)x+2/3 and y=(-4/3)x+10/3. I

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Could someone please help me with writing a conic equation in standard form for a hyperbola with vertices at (1,6) and (1,-2) and asymptotes at y=(4/3)x+2/3 and y=(-4/3)x+10/3. I       Log On


   

Question 1028772: Could someone please help me with writing a conic equation in standard form for a hyperbola with vertices at (1,6) and (1,-2) and asymptotes at y=(4/3)x+2/3 and y=(-4/3)x+10/3. I would appreciate any help!!!
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The transverse axis should be parallel to the y-axis, and the center is at the point (1,2), the midpoint of the vertices (1,6) and (1,-2).
The standard form is then given by %28y-2%29%5E2%2Fa%5E2+-+%28x-1%29%5E2%2Fb%5E2+=+1
<==> y-2 = +/- %28a%2Fb%29sqrt%28b%5E2%2B%28x-1%29%5E2%29
For very large positive/negative x-values,
y-2 = +/-%28a%2Fb%29%28x-1%29%29. (Suppress b%5E2.
==> y = +/-%28a%2Fb%29%28x-1%29%29 + 2.
y = +/-%284%2F3%29%28x-1%29%29 + 2.
Hence let a = 4 and b = 3, and the standard form is %28y-2%29%5E2%2F16+-+%28x-1%29%5E2%2F9+=+1