The transverse axis of a hypserbola is of length 2a and a vertex divides the segment of the axis between the centre and the corresponding focus in the ratio 2 : 1. The equation of the hyperbola is

let hyperbola be $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

$\mathrm{C}=(0, 0), \mathrm{S}=(\mathrm{ae}, 0), \mathrm{A}=(\mathrm{a}, 0)$

Given A divides $\overline{\mathrm{CS}}$ in the ratio 2 : 1

$$\begin{gathered} \Rightarrow (\mathrm{a}, 0)=\left(\frac{2\left(\mathrm{ae}\right)}{3},0\right) \\ \Rightarrow \mathrm{a}=\frac{2\mathrm{ae}}{3} \\ \Rightarrow 2\mathrm{e}=3 \\ \Rightarrow \mathrm{e} = \frac{3}{2} \\ \mathrm{Now} {\mathrm{b}}^2={\mathrm{a}}^2({\mathrm{e}}^2-1) \\ ={\mathrm{a}}^2\left(\frac{5}{4}\right) \\ =\frac{5{\mathrm{a}}^2}{4} \\ \therefore \mathrm{hyperbola} \mathrm{is} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\displaystyle \frac{5{\mathrm{a}}^2}{4}}}=1 \\ \Rightarrow \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{4y^2}{5a^2}=1 \Rightarrow 5{\mathrm{x}}^2-4{\mathrm{y}}^2=5{\mathrm{a}}^2 \end{gathered}$$