SOLUTION: 1. find the vertices and foci of the hyperbola x^2-y^2=1 --- --- 144 9 2.write an equation for the hyperbola. its center is (0,0) foci:(0,-9),(0,9) vertices:(0,-7),

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 1. find the vertices and foci of the hyperbola x^2-y^2=1 --- --- 144 9 2.write an equation for the hyperbola. its center is (0,0) foci:(0,-9),(0,9) vertices:(0,-7),      Log On


   



Question 112893: 1. find the vertices and foci of the hyperbola
x^2-y^2=1
--- ---
144 9


2.write an equation for the hyperbola. its center is (0,0)
foci:(0,-9),(0,9) vertices:(0,-7),(0,7)
if you can help me i would appriciate it thak you

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
find the vertices and foci of the hyperbola
x%5E2%2F144-y%5E2%2F9=1
If a hyperbola opens in the horizontal direction, the equation of a hyperbola in standard form appears as
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1
since a%5E2=144
b%5E2=9
=> a = +-12
b = +-3
then we have
center: (h,k) = (0,0)
focal length:
f=+-+%28sqrt+%28144%2B9+%29%29+
and foci: (0,12.37) (0,-12.37)
the vertices:
since a%5E2=144=> a= +-12, then the vertices will be at:
(-12,0) (12,0)

2.write an equation for the hyperbola. its center is (0,0)
foci:(0,-9),(0,9)
vertices:(0,-7),(0,7)

since center (h,k) is at (0,0),
=> this is a hyperbola opens in the horizontal direction, the equation of a hyperbola in standard form will be
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1

since foci are at (0,-9), (0,9)
=> a = +-9
=>a%5E2=81 and
vertices are at (0,-7),(0,7) =>
b=+%2B-7 =>b%5E2=49
then your equation will be:
x%5E2%2F81-y%5E2%2F49=1