# Equation of a Hyperbola

• Nov 23rd 2009, 04:44 PM
l flipboi l
Equation of a Hyperbola
Hello,

Can someone please verify if I'm correct with setting up the problem?

I'm trying to find the equation for a hyperbola

((x-4)^2 / 2) - (y^2 / 2) = 1

I've attached the graph.

Thanks!
• Nov 23rd 2009, 05:02 PM
apcalculus
Your y-intercepts seem to be incorrect. When you plug in x=0 into your equation, you do not get 4 and -4 for the y-values.

The general equation is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

you can then shift this anyway you like.

More at : Hyperbola -- from Wolfram MathWorld
• Nov 23rd 2009, 08:13 PM
l flipboi l
(oops need to redo)

you're right, then the answer should be:

$(x-2)^2/4 - y^2/2 = 1$
• Nov 23rd 2009, 08:55 PM
l flipboi l
I seem to be stuck...

The equation I'm working with is

$(x-h)^2/a^2 - (y-k)^2/b^2=1$

now when i substitute the given, i get this:

$(x-4)^2/2 - (y-0)^2/b^2=1$

I'm trying to solve for $b^2$

and I keep getting b^2=48

• Nov 24th 2009, 10:50 AM
apcalculus
Quote:

Originally Posted by l flipboi l
I seem to be stuck...

The equation I'm working with is

$(x-h)^2/a^2 - (y-k)^2/b^2=1$

now when i substitute the given, i get this:

$(x-4)^2/2 - (y-0)^2/b^2=1$

I'm trying to solve for $b^2$

and I keep getting b^2=48

Note that

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

has x-intercepts: -a and a
has y-intercepts: -b and b
centered at (0, 0)

Shifting this 5 units to the right, and stretching it vertically, we get:

$\frac{(x-4)^2}{2^2} - \frac{3 y^2}{4^2} = 1$
• Nov 25th 2009, 01:01 PM
l flipboi l
Quote:

Originally Posted by apcalculus
Note that

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

has x-intercepts: -a and a
has y-intercepts: -b and b
centered at (0, 0)

Shifting this 5 units to the right, and stretching it vertically, we get:

$\frac{(x-4)^2}{2^2} - \frac{3 y^2}{4^2} = 1$

Thanks! what's the algebraic step of getting $3y^2/4^2$