Asked by Hal
(-2,1) and (2,1)
Passes through (5,4)
What is the standard form of the equation of the hyperbola?
Passes through (5,4)
What is the standard form of the equation of the hyperbola?
Answers
Answered by
Hal
(-2,1) and (2,1) are the vertices.
Answered by
Reiny
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)
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)
Answered by
Hal
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?
x^2/4 - (y-1)^2/(9/5)=1
because that's what you get plug b^2 back into the original equation?
Answered by
Reiny
5(y-1)^2/9 is the same as (y-1)^2/(9/5)
I don't like double decker fractions
I don't like double decker fractions
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