Question

If the distance between the foci fo a hyperbola is 16 its ecentricity is $\sqrt{2}$.then then obtain its equation.

Answer 1

Eccentricity$e=\sqrt{2}$

Distance between foci is 2ae=16 $2a\sqrt{2}=16$

$a=\frac{16}{23\sqrt{2}}=16$

$a=\frac{16}{23\sqrt{2}}=4\sqrt{2}$

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

$\sqrt{2}=\frac{\sqrt{32+b^2}}{4\sqrt{2}}$

$8=\sqrt{32+b^2}$

$64=32+b^2$

$b^2=32$

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

Rewriting we get,$x^2-y^2=32$