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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equations of the hyperbola satisfying the given conditions : Vertices $(0, \pm 3),$ foci $(0, \pm 5)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{16}-\large\frac{y^2}{9}=1 & \quad (B)\;\large\frac{x^2}{9}-\large\frac{y^2}{16}=1 \\ (C)\;\large\frac{x^2}{16}+\large\frac{y^2}{9}=1 & \quad (D)\;\large\frac{x^2}{9}+\large\frac{y^2}{16}=1 \end {array}$

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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 :
Coordinates of vertices and foci are $(0, \pm 3 )$ and foci $(0, \pm 5)$ respectively.
Clearly the vertices are on the y - axis.
Hence the equation of the hyperbola is of the form $ \large\frac{y^2}{a^2}$$ - \large\frac{x^2}{b^2}$$=1$
Since the vertices are $(0, \pm 3)$
$ \therefore a = 3$
Since the foci are $(0, \pm 5) \Rightarrow c = 5$
$ \therefore c^2 = a^2+b^2$
$ \Rightarrow b^2 = 25-9$
$ = 16$
Hence the equation of the hyperbola is
$ \large\frac{y^2}{9}$$ - \large\frac{x^2}{16}$$=1$
answered Jul 21, 2014 by thanvigandhi_1
 

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