Comment
Share
Q)

# The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt 2$.Its equation is

$\begin{array}{1 1}x^2-y^2=32\\\large\frac{x^2}{4}-\frac{y^2}{9}\normalsize =1\\2x-3y^2=7\\4x^2-y^2=35\end{array}$

• General equation of a hyperbola which lies on the $y$ axis is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1 • a^2=b^2(e^2-1) Answer : x^2-y^2=32 The distance between the foci =16 (ie) 2ae=16 \Rightarrow ae=8 Given eccentricity 'e'=\sqrt 2 \therefore a\sqrt 2=8 a=\large\frac{8}{\sqrt 2} a^2=\large\frac{64}{2}$$=32$
$b^2=a^2(e^2-1)$
$\;\;\;\;=32[(\sqrt 2)^2-1]$
$b^2=32$
Hence the equation of the hyperbola is $\large\frac{x^2}{32}-\frac{y^2}{32}$$=1$
Or $x^2-y^2=32$