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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 $\; 5y^2-9x^2=36$

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Given $\;5y^2-9x^2=36. $ Dividing by $36 \rightarrow \large\frac{5}{36}$$y^2-\large\frac{9}{36}$$x^2=1 \rightarrow \large\frac{y^2}{\frac{6}{\sqrt5}} $$-\large\frac{x^2}{2}$$=1$
On comparing the given equation with the standard equation of a hyperbola, we get:
$\quad a =\large\frac{6}{\sqrt 5}$$, b = 2 $
Since $a^2+b^2 = c^2 \rightarrow c = \sqrt{\frac{36}{5}+4} $$= \sqrt {\large\frac{56}{5}}$ $ = 2\large\sqrt{\frac{14}{5}}$
1. Coordinates of Foci $ = (0, \pm 2\large\sqrt{\frac{14}{5}}$$)$
2. Coordinates of Vertices $= (0,\pm \large\frac{6}{\sqrt 5}$$)$
3. Eccentricity $e = \large\frac{c}{a} $$= 2\large\sqrt{\frac{14}{5}} $$ \times \large\frac{\sqrt 5}{6}$$ = \large \frac{\sqrt{14}}{3}$
4. Latus Rectum $=\large\frac{2b^2}{a}$$ = \large\frac{2\times 2^2}{\frac{6}{\sqrt 5}} $$= 4\large\frac{\sqrt 5}{3}$
answered Apr 8, 2014 by balaji.thirumalai
edited Apr 8, 2014 by balaji.thirumalai
 

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