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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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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 $= (0,\pm 2)$
3. Eccentricity $e = \large\frac{c}{a} $$=\large\frac{\sqrt{13}}{2}$
4. Latus Rectum $=\large\frac{2b^2}{a}$$ = \large\frac{2\times 3^2}{2} $$= 9$
answered Apr 8, 2014 by balaji.thirumalai
edited Apr 8, 2014 by balaji.thirumalai
 

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