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

Find the equation of the hyperbola with vertices $(0,\pm 7),e=\large\frac{4}{3}$.

$\begin{array}{1 1}\large\frac{9x^2}{343}-\frac{y^2}{49}\normalsize =1\\\large\frac{9x^2}{343}+\frac{y^2}{49}\normalsize =1\\\large\frac{9x^2}{343}+\frac{y^2}{49}\normalsize =-1\\\large\frac{9x^2}{343}-\frac{y^2}{49}\normalsize =-1\end{array} $

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A)
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  • If the vertices lie on the $y$ axis then the equation of the hyperbola is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1$
Answer : $\large\frac{9x^2}{343}-\frac{y^2}{49}\normalsize =-1$
Given data are the vertices are $(0,\pm 7)$ and $e=\large\frac{4}{3}$.
Since the vertices lie on the $y$-axis,the equation of the hyperbola is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1$
The coordinates of this hyperbola are $(0,\pm b)$
$\therefore b=7$
$a^2=b^2(e^2-1)$
(ie) $a^2=49(\large\frac{16}{9}$$-1)=49(\large\frac{16-9}{9})$
$\therefore a^2=\large\frac{343}{9}$
Hence the equation of the required hyperbola is $\large\frac{9x^2}{343}-\frac{y^2}{49}\normalsize =-1$
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