Eccentricity of given hyperbola $=\frac{5}{3}$
Let, eccentricity of its conjugate hyperbola $=e$
So, $\frac{1}{\left(\frac{5}{3}\right)^{2}}+\frac{1}{e^{2}}=1 \begin{bmatrix}\because \text { relationship between } \\ \text { eccentricity is } \frac{1}{ e _{1}^{2}}+\frac{ l }{e_{2}^{2}}=1\end{bmatrix}$
$\Rightarrow \frac{9}{25}+\frac{1}{e^{2}}=1 $
$\Rightarrow \frac{1}{e^{2}}=1-\frac{9}{25}$
$\Rightarrow \frac{1}{e^{2}}=\frac{16}{25} $
$\Rightarrow e^{2}=\frac{25}{16}$
$\Rightarrow e=\frac{5}{4}$
Hence, eccentricity of conjugate hyperbola $ =\frac{5}{4}$