Given:

The condition is \[5{{y}^{2}}~\text{ }9{{x}^{2}}~=\text{ }36\]

Let us divide the whole equation by \[36,\]we get

\[\begin{array}{*{35}{l}}

5{{y}^{2}}/36\text{ }\text{ }9{{x}^{2}}/36\text{ }=\text{ }36/36  \\

{{y}^{2}}/\left( 36/5 \right)\text{ }\text{ }{{x}^{2}}/4\text{ }=\text{ }1  \\

\end{array}\]

On contrasting this condition and the standard condition of hyperbola\[{{y}^{2}}/{{a}^{2}}~\text{ }{{x}^{2}}/{{b}^{2}}~=\text{ }1\],

We get \[a\text{ }=\text{ }6/\surd 5\text{ }and\text{ }b\text{ }=\text{ }2,\]

It is realized that, \[{{a}^{2}}~+\text{ }{{b}^{2}}~=\text{ }{{c}^{2}}\]

So,

\[\begin{array}{*{35}{l}}

{{c}^{2}}~=\text{ }36/5~+\text{ }4  \\

{{c}^{2}}~=\text{ }56/5  \\

c\text{ }=~\surd \left( 56/5 \right)  \\

=\text{ }2\surd 14/\surd 5  \\

Then,  \\

The\text{ }coordinates\text{ }of\text{ }the\text{ }foci\text{ }are\text{ }\left( 0,\text{ }2\surd 14/\surd 5 \right)\text{ }and\text{ }\left( 0,\text{ }\text{ }2\surd 14/\surd 5 \right).  \\

The\text{ }coordinates\text{ }of\text{ }the\text{ }vertices\text{ }are\text{ }\left( 0,\text{ }6/\surd 5 \right)\text{ }and\text{ }\left( 0,\text{ }-6/\surd 5 \right).  \\

Eccentricity,\text{ }e\text{ }=\text{ }c/a\text{ }=\text{ }\left( 2\surd 14/\surd 5 \right)\text{ }/\text{ }\left( 6/\surd 5 \right)\text{ }=~\surd 14/3  \\

Length\text{ }of\text{ }latus\text{ }rectum\text{ }=\text{ }2{{b}^{2}}/a\text{ }=\text{ }\left( 2\text{ }\times \text{ }{{2}^{2}} \right)/6/\surd 5\text{ }=\text{ }\left( 2\times 4 \right)/6/\surd 5\text{ }=\text{ }4\surd 5/3  \\

\end{array}\]