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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 $(0, \pm 5),$ foci $(0, \pm 8)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{39}+\large\frac{y^2}{25}=1 \\ (B)\;\large\frac{x^2}{39}-\large\frac{y^2}{25}=1 \\ (C)\;\large\frac{x^2}{25}+\large\frac{y^2}{39}=1 \\ (D)\;\large\frac{x^2}{25}-\large\frac{y^2}{39}=1 \end {array}$

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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 coordinates of vertices and foci given are $( 0, \pm 5 )$ and $(0, \pm 8 )$ respectively.
Clearly the vertices lie on the y - axis.
Hence the equation should be of the form $ \large\frac{y^2}{a^2}$$ - \large\frac{x^2}{a^2}$$=1$
Since the vertices are $( 0, \pm 5 )$ a = 5
Since the foci are $(0, \pm 8)$, c = 8
$ \therefore c^2 = a^2+b^2$
Substituting the values for a and c we get,
$ 5^2+b^2=8^2$
$ \therefore b^2 = 64-25 = 39$
The equation of the hyperbola is
$ \large\frac{y^2}{25}$$ - \large\frac{x^2}{39}$$=1$
answered Jul 21, 2014 by thanvigandhi_1
 

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