If the equation of the hyperbola is $9 y^{2} - 4 x^{2} = 36$ , then vertices and length of latus rectum is

(0, + 2), 0

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(0, - 2), 0

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(0, \pm 2), 9

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N o n e o f t h e s e

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Solution

The correct option is C $(0, \pm 2), 9$
$9 y^{2} - 4 x^{2} = 36 \Rightarrow \frac{9 y^{2}}{36} - \frac{4 x^{2}}{36} = 1 \frac{y^{2}}{4} - \frac{x^{2}}{9} = 1$
Comparing $= \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$
We have $b^{2} = 4, b = 2 a^{2} = 9, a = 3$
Co-ordinate of vertex $= (0, \pm 5) = (0, \pm 2)$
Length of the Lotus rectum $= \frac{2 a^{2}}{b} = \frac{2 \times 9}{2} = 9 u n i t s$

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9 y^{2} - 4 x^{2} = 36

Find the coordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of the hyperbola,

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

9 y^{2} - 4 x^{2} = 36

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