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

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt 2$.Its equation is

$\begin{array}{1 1}x^2-y^2=32\\\large\frac{x^2}{4}-\frac{y^2}{9}\normalsize =1\\2x-3y^2=7\\4x^2-y^2=35\end{array} $

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A)
Toolbox:
  • General equation of a hyperbola which lies on the $y$ axis is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1$
  • $a^2=b^2(e^2-1)$
Answer : $x^2-y^2=32$
The distance between the foci $=16$
(ie) $2ae=16$
$\Rightarrow ae=8$
Given eccentricity $'e'=\sqrt 2$
$\therefore a\sqrt 2=8$
$a=\large\frac{8}{\sqrt 2}$
$a^2=\large\frac{64}{2}$$=32$
$b^2=a^2(e^2-1)$
$\;\;\;\;=32[(\sqrt 2)^2-1]$
$b^2=32$
Hence the equation of the hyperbola is $\large\frac{x^2}{32}-\frac{y^2}{32}$$=1$
Or $ x^2-y^2=32$
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