Question

Find the equation of hyperbola whose eccentricity is $\sqrt{2}$ and distance between foci is 16. Also name the hyperbola so obtained.

Answer 1

Given $e=\sqrt{2}$ for hyperbola and distance between foci is $16$ and we know distance between foci $=2C=16$

$[C=8]$

and $C=\sqrt{a^2+b^2}=8$ ……….$\left(1\right)$

and $e=\sqrt{2}$

We know $a=\frac{C}{B}=\frac{8}{\sqrt{2}}=4\sqrt{2}$

$[a=4\sqrt{2}]$

and from equation $\left(1\right)$ we get

$\sqrt{a^2+b^2}=8$

Squaring both sides we get

$a^2+b^2=64$

$b^2=64-32$

$b^2=32$

$[b=4\sqrt{2}]$

Hence equation of hyperbola is $\frac{x^2}{32}-\frac{y^2}{32}=1$.