The equation of the hyperbola whose centre is (1, 2), one focus is (6, 2) and transverse axis 6 is

$$\begin{gathered} \mathrm{Given} \mathrm{C}=(1, 2), \mathrm{focus}\left(\mathrm{S}\right)=(6, 2) \\ \mathrm{and} length of trasverse axis=2\mathrm{a}=6 \Rightarrow \mathrm{a}=3 \\ \mathrm{Now} \mathrm{CS}=\sqrt{25}=5 \\ \Rightarrow ae=5 \\ \Rightarrow \mathrm{e}=\frac{5}{3} \\ \mathrm{Now} {\mathrm{b}}^2={\mathrm{a}}^2({\mathrm{e}}^2-1) \\ =9(\frac{25}{9}-1) \\ =9\left(\frac{16}{9}\right) \\ =16 \\ \therefore \mathrm{Hyperbola} \mathrm{is} \frac{{(\mathrm{x}-1)}^2}{9}-\frac{{(\mathrm{y}-2)}^2}{16}=1 \Rightarrow 16{(\mathrm{x}-1)}^2-9{(\mathrm{y}-2)}^2=144 \end{gathered}$$