For finding the shape of the hyperbola, trace the equation $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

I have no clue what to do???

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- Jan 6th 2010, 01:12 AMTomJerrySimple problem but no clue
For finding the shape of the hyperbola, trace the equation $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

I have no clue what to do??? - Jan 6th 2010, 02:15 AMmr fantastic
- Jan 6th 2010, 02:22 AMabender$\displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 $

whereby

$\displaystyle c^2 = a^2 + b^2 $ .

In your hyperbola, (h,k) = (0,0), so we have a start to this: The center is at the origin.

Note that if

$\displaystyle \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 $

then the transverse axis would be vertical. However, the way we have it, with the x-term positive and the y-term negative, the traverse axis is horizontal