Question 757927
9x^2-16y^2=144
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divide 144 to the two sides we get:
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x^2/16 - y^2/9 = 1
<br>
the equation can be written as:
<br>
(x-0)^2/16 - (y-0)^2/9 = 1
<br>
we get:
h=0
k=0
a^2 = 16 -> a = 4
b^2 = 9 -> b = 3
<br>
now, we must find c
c^2 = a^2 + b^2
c^2 = 16 + 9
c^2 = 25
c = 5
<br>
foci:
(h±c,k)
(0±5,0)
(-5,0) and (5,0)
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vertices:
(h±a,k)
(0±4,0)
(-4,0) and (4,0)
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center
(h,k)
(0,0)
<br>
asymptotes
y=±b/a(x-h)+k
y=±3/4(x-0)+0
y=±3x/4
<br>
it's hyperbola