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# Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. 9 y ^ 2 - 4 x ^ 2 = 36

3.  Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

$9 y^2 - 4 x^2 =36$

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Given a Hyperbola equation,

$9 y^2 - 4 x^2 =36$

Can also be written as

$\frac{9y^2}{36} - \frac{4x^2}{36} = 1$

$\frac{y^2}{2^2} - \frac{x^2}{3^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

We get,

$a=2$ and $b=3$

Now, As we know the relation in a hyperbola,

$c^2=a^2+b^2$

$c=\sqrt{a^2+b^2}$

$c=\sqrt{2^2+3^2}$

$c=\sqrt{13}$

Hence,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=(0,\sqrt{13})\:and\:(0,-\sqrt{13})$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=(0,2)\:and\:(0,-2)$

The Eccentricity:

$e=\frac{c}{a}=\frac{\sqrt{13}}{2}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(9)}{2}=\frac{18}{2}=9$

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