Find the eq of Hyperbola whose eccentricity is $\displaystyle \sqrt{2}$ and the distance between the focii is 16.

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- Dec 29th 2008, 11:52 PMvarunnayuduFind eq of hyperbola
Find the eq of Hyperbola whose eccentricity is $\displaystyle \sqrt{2}$ and the distance between the focii is 16.

- Dec 30th 2008, 12:38 AMnzmathman
For a hyperbola, $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,

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

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

$\displaystyle 2a = 16 $

$\displaystyle \therefore a = 8$

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

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

$\displaystyle \Rightarrow 64 + b^2 = 128$

$\displaystyle \Rightarrow b^2 = 64$

$\displaystyle \Rightarrow b = 8$

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

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

i need to check ..........(Smirk) - Dec 30th 2008, 01:16 AMmr fantastic