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# Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $\;9y^2-4x^2=36$

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Given $\;9y^2-4x^2=36.$ Dividing by $36 \rightarrow \large\frac{9}{36}$$y^2-\large\frac{4}{36}$$x^2=1 \rightarrow \large\frac{y^2}{4} $$-\large\frac{x^2}{9}$$=1$
On comparing the given equation with the standard equation of a hyperbola, we get:
$\quad a =2, b = 3$
Since $a^2+b^2 = c^2 \rightarrow c = \sqrt{4+9} = \sqrt {13}$
1. Coordinates of Foci $= (0, \pm \sqrt{13})$
2. Coordinates of Vertices $= (0,\pm 2)$

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