The distance between the foci of a hyperbola is 16 and its eccentricity is `sqrt(2)` then equation of the hyperbola is

Explanation

(i) Let the equation of the hyperbola be `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`.
Foci are `(pmae,0)`.
Distance between foci = 2ae = 16 (given)
Also, `e=sqrt2` (given)
`therefore" "a=4sqrt2`
We know that,
`b^(2)=a^(2)(e^(2)-1)`
`therefore" "b^(2)=(4sqrt2)^(2)[(sqrt2)^(2)-1]=32`
So, the equation of hyperbola is
`(x^(2))/(32)-(y^(2))/(32)=1`
or `x^(2)-y^(2)=32`
If the hyperbola is of the form `(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1`, then its equation will be `x^(2)-y^(2)=-32`
(ii) Vertices are `(pm4,0)` and foci are `(pm6,0).`
`therefore" "a=4 and ae=6`
`rArr" "e=(6)/(4)=(3)/(2)`
Now, `b^(2)=a^(2)(e^(2)-1)`
`therefore" "b^(2)=16((9)/(4)-1)=20`
So, the equation of hyperbola is `(x^(2))/(16)-(y^(2))/(20)=1.`
(iii) Since foci `(0,pmsqrt(10))` are on y-aixs, consider the equation of hyperbola as `(x^(2))/(y^(2))-(y^(2))/(b^(2))=-1.`
Also, `a^(2)=b^(2)(e^(2)-1)`
`therefore" "a^(2)=b^(2)e^(2)-b^(2)=10-b^(2)`
`therefore" Equation of the hyperbola becomes"`
`(x^(2))/(10-b^(2))-(y^(2))/(b^(2))=-1`
Since, hyperbola passes through the point (2, 3), we have
`therefore" "(4)/(10-b^(2))-(9)/(b^(2))=-1`
`rArr" "4b^(2)-9(10-b^(2))=-b^(2)(10-b^(2))`
`rArr" "b^(4)-23b^(2)+90=0`
`rArr" "(b^(2)-18)(b^(2)-5)=0`
`rArr" "b^(2)=5(b^(2)=18 " not possible as " a^(2)+b^(2)=10)`
`rArr" "a^(2)=10-5=5`
So, the equation of hyperbola is
`(x^(2))/(5)-(y^(2))/(5)=-1`
or `y^(2)-x^(2)=5`
(iv) According to the equation
`ae-a=1 and ae+a=3`
`therefore" "(e+1)/(e-1)=3 or e=2`
`therefore" "a=1`
So, `b^(2)-a^(2)(e^(2)-1)=3`
Therefore, equation is `x^(2)-(y^(2))/(3)=1` or it can be `(x^(2))/(3)-y^(2)=-1`.