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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 hyperbolas : $5y^2-9x^2=36$

$\begin {array} {1 1} (A)\;Foci : ( 0, \pm 2\sqrt{14} / \sqrt{5} ), Vertices : ( 0, \pm 6 / \sqrt 5), e = \sqrt{14} / 3, Length \; of \: the \: latus\: rectum 4\sqrt 5 / 3 \\ (B)\;Foci : ( \pm 2\sqrt{14} / \sqrt{5} , 0), Vertices : ( \pm 6 / \sqrt 5, 0), e = \sqrt{14} / 3, Length \; of \: the \: latus\: rectum 4\sqrt 5 / 3 \\ (C)\;Foci : ( 2\sqrt{14} / 5, 0 ), Vertices : ( 6 / \sqrt 5, 0), e = \sqrt{14} / 3, Length \; of \: the \: latus\: rectum 4\sqrt 5 / 3 \\ (D)\;Foci : ( -\pm 2\sqrt{14} / \sqrt{5}, 0 ), Vertices : ( -6 / \sqrt 5, 0), e = \sqrt{14} / 3, Length \; of \: the \: latus\: rectum 4\sqrt 5 / 3 \end {array}$

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Toolbox:
  • Standard equation of a hyperbola where transverse axis is along x - axis is $ \large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2} $$=1$
  • Standard equation of a hyperbola whose transverse axis is along y - axis is $ \large\frac{y^2}{a^2} $$ - \large\frac{x^2}{b^2}$$=1$
  • $c^2=a^2+b^2$, where $c$ is the foci
  • Eccentricity $e = \large\frac{c}{a}$
  • Length of the latus rectum $ = \large\frac{2b^2}{a}$
Step 1 :
Given equation is $5y^2-9x^2=36$
Dividing throughout by 36 we get,
$ \large\frac{y^2}{\Large\frac{36}{5}} $$ - \large\frac{x^2}{4}$$=1$
Comparing this equation with the standard equation of the hyperbola
$\large\frac{y^2}{a^2}$$- \large\frac{x^2}{b^2}$$=1$ we get,
$a^2= \large\frac{36}{5}$ and $ b^2=4$
$ \therefore c^2 = a^2+b^2$
(i.e) $ c^2= \large\frac{36}{5}$$+4 = \large\frac{56}{5}$
$ \therefore c = \sqrt{ \large\frac{56}{5}}$$ = \large\frac{2\sqrt{14}}{\sqrt 5}$
Step 2 :
Eccentricity $ e = \large\frac{c}{a}$$=\large\frac{\sqrt{ \large\frac{56}{5}}}{\large\frac{6}{\sqrt 5}}$
$ = \large\frac{\sqrt{14}}{3}$
Coordinates of foci are $ \bigg(0, \pm \large\frac{2\sqrt{14}}{\sqrt 5} \bigg)$
Coordinates of vertices are $ \bigg(0, \pm \large\frac{6}{\sqrt 5} \bigg)$
Length of the latus rectum $ = \large\frac{2b^2}{a}$$ = \large\frac{2 \times 4 }{\Large\frac{6}{\sqrt 5}}$
$ = \large\frac{4\sqrt 5 }{3}$
answered Jul 19, 2014 by thanvigandhi_1
 

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