The equation is 5y^{2} – 9x^{2} = 36

Let us divide the whole equation by 36, we get

\( \dfrac{5y^2}{9}- \dfrac{9x^2}{36}=1\)

= \( \dfrac{36}{36}\)

\( \dfrac{y^2}{\dfrac{36}{5}}-\dfrac{x^2}{4}=1\)

On comparing this equation with the standard equation of hyperbola

\( \dfrac{x^2}{a^2}\) +\( \dfrac{y^2}{b^2}\) = 1,

We get a = \( \dfrac{6}{\sqrt{5}}\) and b = 2,

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

So,

c^{2} = \( \dfrac{36}{5}\) + 4

c^{2} = \( \dfrac{56}{5}\)

c = \( \sqrt{\dfrac{56}{5}}\)

= \( \dfrac{2\sqrt{14}}{\sqrt{5}}\)

Then,

The coordinates of the foci are

(0, \( \dfrac{2\sqrt{14}}{\sqrt{5}}\)) and (0, – \( \dfrac{2\sqrt{14}}{\sqrt{5}}\)).

The coordinates of the vertices are

(0,\( \dfrac{6}{\sqrt{5}}\)) and (0, \( - \dfrac{6}{\sqrt{5}}\)).

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

= \( \dfrac{ \dfrac{2\sqrt{14}}{\sqrt{5}}}{\dfrac{6}{\sqrt{5}}}\)

= \(\dfrac{\sqrt{14}}{3}\)

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

= \( \dfrac{(2×2^2)}{\dfrac{6}{\sqrt{5}}}\)

= \( \dfrac{(2×4)}{\dfrac{6}{\sqrt{5}}}\)

= \(\dfrac{4\sqrt{5}}{3}\)

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