Question 407514: Find the center, vertices, foci and axis of semmetry, then graph the equation.
8x^2-3y^2=48
thank you...
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the center, vertices, foci and axis of semmetry, then graph the equation.
8x^2-3y^2=48
..
Standard forms of hyperbola:
(x-h)^2/a-(y-k)^2=1 (horizontal transverse axis)
(y-k)^2/a-(x-h)^2=1 (vertical transverse axis)
8x^2-3y^2=48
x^2/6-y^2/16=1
This is a hyperbola with horizontal transverse axis. (opens sideways)
a^2=6
a=sqrt(6)=2.45
b^2=16
b=4
c=sqrt(a^2+b^2)=sqrt(6+16)=sqr(22)=4.69
center: (0,0)
vertex: (0+-a,0)=(-2.45,0),(2.45,0)
foci: (0+-c,0)=(-4.69,0),(4.69,0)
Transverse axis: y=0 (this is what I believed you called the axis of symmetry, which normally applies to parabolas)
The graph should look similar to the graph below
..
y=(8x^2/3)-16)^.5
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