Correct Answer - A
Given that, distance between the foci of hyperbola
i.e., 2ae=16 `rArrae=8`….(i)
and `e=sqrt2`…(ii)
Now, `sqrt2a=8`
`rArr a=4sqrt2`
`becauseb^(2)=a^(2)(e^(2)-1)`
`rArr b^(2)=32(2-1)`
`rArrb^(2)=32`
`therefore (x^(2))/32-(y^(2))/32=1`
`rArrx^(2)-y^(2)=32`