Question
(-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
Hal
(-2,1) and (2,1) are the vertices.
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)
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?
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