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

$\begin {array} {1 1} (A)\;Foci ( 0, \pm 6), Vertices (0, \pm 3), e = 2, Length \: of \: the \: latus \: rectum : 18 \\ (B)\;Foci ( \pm 6, 0), Vertices ( \pm 3, 0), e = 2, Length \: of \: the \: latus \: rectum : 18 \\ (C)\;Foci ( 0, \pm 6), Vertices (0, \pm 3), e = 1, Length \: of \: the \: latus \: rectum : 18 \\ (D)\;Foci ( 0, \pm 6), Vertices (0, \pm 3), e = 2, Length \: of \: the \: latus \: rectum : 2 \end {array}$

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  • 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 :
The given equation is $ \large\frac{y^2}{9}$$-\large\frac{x^2}{27}$$=1$
Comparing this with the standard equation of a hyperbola, $ \large\frac{y^2}{a^2}$$-\large\frac{x^2}{b^2}$$=1$ we get,
$a^2=9 \Rightarrow a = 3$
$ b^2=27 \Rightarrow b = 3\sqrt 3$
$ \therefore c^2 = a^2 + b^2$
(i.e) $c^2 = 9+ 27=36$
$ \Rightarrow c = \pm 6$
Step 2 :
The coordinates of foci are $(0, \pm 6)$
Coordinates of vertices are $( 0, \pm 3)$
Eccentricity $e = \large\frac{6}{3}$$=2$
Length of the latus rectum = $ \large\frac{2b^2}{a}$$=\large\frac{ 2 \times 27}{3}$
$ = 18$
answered Jul 19, 2014 by thanvigandhi_1
 

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