Find the equation of the hyperbola, whose foci are (4, 2) and (8, 2) and eccentricity is 2.

Given foci are $\mathrm{S}=(4,2),{\mathrm{S}}^1=(8,2) and \mathrm{e}=2$
$\mathrm{Since} \mathrm{y}-{\text{ co-ordinates of S and S}}^1\text{ are same then}$$\text{ equation of transverse axis is }{\parallel }^{le}\text{ to }$x-axis
$\therefore$Equation of hyperbola is $\frac{(\mathrm{x}-\mathrm{h}{)}^2}{{\mathrm{a}}^2}-\frac{(\mathrm{y}-\mathrm{k}{)}^2}{{\mathrm{b}}^2}=1$
$\therefore$centre C(h, k) = midpoint of S and S¹
                              $=\left(\frac{4+8}{2},\frac{2+2}{2}\right)=(6,2)$
Distance between foci $2\mathrm{ae}=S{S}^1$
$\begin{array}{l}2\mathrm{ae}=\sqrt{(4-8{)}^2+(2-2{)}^2}\\ 2\mathrm{ae}=4\Rightarrow \mathrm{ae}=2\Rightarrow \mathrm{a}\left(2\right)=2(\because \mathrm{e}=2)\end{array}$
$\begin{array}{l}\mathrm{a}=1\Rightarrow {\mathrm{a}}^2=1\\ \Rightarrow {\mathrm{b}}^2={\mathrm{a}}^2\left({\mathrm{e}}^2-1\right)=1\left(2^2-1\right)=3\end{array}$
$\therefore \text{ Equation of hyperbola is }$$\frac{(\mathrm{x}-6{)}^2}{1}-\frac{(\mathrm{y}-2{)}^2}{3}=1$