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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 $\;\large\frac{y^2}{9}$$-\large\frac{x^2}{27}$$=1$

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On comparing the given equation with the standard equation of a hyperbola, we get:
$\quad a = \sqrt{9} = 3, b = \sqrt{27} $
Since $a^2+b^2 = c^2 \rightarrow c = \sqrt{9+27} = \sqrt {36} = 6$
1. Coordinates of Foci $ = (0, \pm 6)$
2. Coordinates of Vertices $= (0,\pm 3)$
3. Eccentricity $e = \large\frac{c}{a} $$=\large\frac{6}{3}$$=2$
4. Latus Rectum $=\large\frac{2b^2}{a}$$ = \large\frac{2\times\sqrt{27}^2}{3} $$= 18$
answered Apr 8, 2014 by balaji.thirumalai
 

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