Saturday
May 18, 2013

# Homework Help: Pre-Cal

Posted by Hal on Saturday, May 2, 2009 at 7:37pm.

(-2,1) and (2,1)
Passes through (5,4)

What is the standard form of the equation of the hyperbola?

• Pre-Cal - Hal, Saturday, May 2, 2009 at 7:38pm

(-2,1) and (2,1) are the vertices.

• Pre-Cal - Reiny, Saturday, May 2, 2009 at 7:55pm

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)

• Pre-Cal - Hal, Saturday, May 2, 2009 at 7:59pm

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?

• Pre-Cal - Reiny, Saturday, May 2, 2009 at 8:15pm

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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