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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

The equation of the hyperbola with vertices at $(0,\pm 6)$ and eccentricity $\large\frac{5}{3}$ is _________ and its foci are ___________

$\begin{array}{1 1}\large\frac{y^2}{36}-\frac{x^2}{64}\normalsize =1,(0,\pm 10)\\\large\frac{y^2}{36}+\frac{x^2}{64}\normalsize =-1,(0,\pm 12)\\\large\frac{x^2}{36}-\frac{y^2}{64}\normalsize =-1,(0,\pm 20)\\\large\frac{y^2}{64}-\frac{x^2}{36}\normalsize =1,(0,\pm 15)\end{array} $

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A)
Toolbox:
  • General equation of a hyperbola with its vertices on $y$ axis is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1$
  • $a^2=b^2(e^2-1)$
The given vertices are $(0,\pm 6)$ and eccentricity $e=\large\frac{5}{3}$
$b=6$
Foci $be =\large\frac{5}{3}$$\times 6=10$
Hence the foci of the hyperbola are $(0,\pm 10)$
$a^2=b^2(e^2-1)$
$\Rightarrow a^2=36(\large\frac{25}{9}$$-1)$
$\Rightarrow 36\times \large\frac{16}{9}$$=64$
Hence the equation of the hyperbola is $\large\frac{y^2}{36}-\frac{x^2}{64}$$=1$
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