Vertical Shift of Hyperbola

**Sw0rDz** · November 4th 2009, 03:55 PM

Question: Find formula for hyperbola with given information.

foci (2,0), (2,8)
asymptotes: y = 3 + x/2 and y = 5 - x/2

My current equation is:
$\frac{(y-4)^2}{2^2} - \frac{(x-2)^2}{1^2} = 1$

My graph:

I'm trying to figure out the last vertical shift.

**earboth** · November 5th 2009, 12:45 AM

Originally Posted by Sw0rDz

Question: Find formula for hyperbola with given information.

foci (2,0), (2,8)
asymptotes: y = 3 + x/2 and y = 5 - x/2

My current equation is:
$\frac{(y-4)^2}{2^2} - \frac{(x-2)^2}{1^2} = 1$

...

According to the given information you know:

Midpoint M(2, 4)

$\dfrac ab =\dfrac12~\implies~a=2b$

Distance of one focus to M is 4; therefore:

$a^2+b^2=16~\implies~4b^2+b^2=16$

Calculate the lengthes of a and b and determine the equation of the hyperbola.

I've got:

$\dfrac{5(y-4)^2}{4^2}-\dfrac{5(x-2)^2}{8^2}=1$

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