Posted by Hal on Saturday, May 2, 2009 at 7:37pm.
(-2,1) and (2,1) are the vertices.
the centre is clearly (0,1), the midpoint of these two vertices.
and a = 2
so x^2/4 - (y-1)^2/b^2 = 1
but it passes through the point (5,4), so
25/4 - 9/b^2 = 1
-9/b^2= 1-24/4
-9/b^2 = -20/4
b^2/9 = 4/20 = 1/5
b^2 = 9/5
so the hyperbola is
x^2/25 - 5(y-1)^2/9 = 1
(check my arithmetic)
Wouldn't it be:
x^2/4 - (y-1)^2/(9/5)=1
because that's what you get plug b^2 back into the original equation?
5(y-1)^2/9 is the same as (y-1)^2/(9/5)
I don't like double decker fractions
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