Given:

The equation is 9y^{2} – 4x^{2} = 36

or \(\dfrac{ y^2}{4} – \dfrac{x^2}{9} = 1\)

or \(\dfrac{ y^2}{2^2} –\dfrac{ x^2}{3^2} = 1\)

On comparing this equation with the standard equation of hyperbola \( \dfrac{ y^2}{a^2} – \dfrac{x^2}{b^2} = 1,\)

We get a = 2 and b = 3,

It is known that, a^{2} + b^{2} = c^{2}

So,

c^{2} = 4 + 9

c^{2} = 13

c = \( \sqrt{13}\)

Then,

The coordinates of the foci are \( (0, \sqrt{13})\) and \( (0, -\sqrt{13})\)

The coordinates of the vertices are (0, 2) and (0, – 2).

Eccentricity, e = \( \dfrac{c}{a}\) = \( \sqrt{\dfrac{13}{2}}\)

Length of latus rectum = \( \dfrac{2b^2}{a}\)= \(\dfrac{ (2 × 3^2)}{2}\)

= \(\dfrac{ (2×9)}{2} = \dfrac{18}{2}\) = 9

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