Question 407514
Find the center, vertices, foci and axis of semmetry, then graph the equation.
8x^2-3y^2=48
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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

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y=(8x^2/3)-16)^.5
{{{ graph( 300, 200, -5, 5, -10, 10, (8x^2/3-16)^.5,-(8x^2/3-16)^.5) }}}