For ellipse $$ e_1=\sqrt{1-\frac{b^2}{a^2}}=\frac{3}{5} $$ for hyperbola $$ e_2=\frac{5}{3} $$ Let hyperbola be $$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $$ it passes through $(3,0) \Rightarrow \frac{9}{a^2}=1$ $$ \begin{aligned} \Rightarrow \quad & a^2=9 \\ \Rightarrow \quad & b^2=a^2\left(e^2-1\right) \\ & =9\left(\frac{25}{9}-1\right)=16 \end{aligned} $$ ∴ $\quad$ Hyperbola is $$ \frac{x^2}{9}-\frac{y^2}{16}=1 $$