# Find eq of hyperbola

• Dec 29th 2008, 11:52 PM
varunnayudu
Find eq of hyperbola
Find the eq of Hyperbola whose eccentricity is $\sqrt{2}$ and the distance between the focii is 16.
• Dec 30th 2008, 12:38 AM
nzmathman
For a hyperbola, $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,

eccentricity = $\frac{\sqrt{a^2 + b^2}}{a} = \sqrt{2}$

the foci are at a and -a, so distance between the two foci is 2a.

$2a = 16$

$\therefore a = 8$

$\Rightarrow \frac{\sqrt{64 + b^2}}{8} = \sqrt{2}$

$\Rightarrow \sqrt{64 + b^2} = 8\sqrt{2}$

$\Rightarrow 64 + b^2 = 128$

$\Rightarrow b^2 = 64$

$\Rightarrow b = 8$

So now you know $a = b = 8$, simply put these values into the standard equation for a parabola.

Actually, technically speaking the answer should be $\frac{(x-f)^2}{a^2} - \frac{(y-g)^2}{b^2} = 1$ where (f,g) is the centre, because any translation will not affect the eccentricity or distance between the foci.
• Dec 30th 2008, 01:13 AM
varunnayudu
what is the eq..
what is the eq ............(Happy)

i need to check ..........(Smirk)
• Dec 30th 2008, 01:16 AM
mr fantastic
Quote:

Originally Posted by varunnayudu
what is the eq ............(Happy)

i need to check ..........(Smirk)