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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 $(\pm 2, 0),$ foci $(\pm 3, 0)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{4}+\large\frac{y^2}{5}=1 \\ (B)\;\large\frac{x^2}{5}+\large\frac{y^2}{4}=1 \\ (C)\;\large\frac{x^2}{4}-\large\frac{y^2}{5}=1 \\ (D)\;\large\frac{y^2}{4}-\large\frac{x^2}{5}=1 \end {array}$

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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 :
The coordinate of the foci and vertices are $( \pm 3, 0)$ and $(\pm 2, 0)$ respectively.
Clearly the vertices lie on the x - axis.
Hence the equation of the hyperbola should be of the form $ \large\frac{x^2}{a^2}$$ - \large\frac{y^2}{b^2}$$=1$
$ \therefore c^2 = a^2+b^2$ where $c= 3$ and $a = 2$
$ b^2 = 9-4=5$
Hence the equation of the hyperbola is $ \large\frac{x^2}{4}$$ - \large\frac{y^2}{5}$$=1$
answered Jul 19, 2014 by thanvigandhi_1
edited Jul 21, 2014 by thanvigandhi_1
 

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