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

Find the equation of the hyperbola with vertices $(\pm 5,0)$,foci $(\pm 7,0)$.

$\begin{array}{1 1}\large\frac{x^2}{25}+\frac{y^2}{24}\normalsize =1\\\large\frac{x^2}{25}-\frac{y^2}{24}\normalsize =1\\\large\frac{y^2}{25}+\frac{x^2}{24}\normalsize =1\\\large\frac{y^2}{24}+\frac{x^2}{25}\normalsize =1\end{array} $

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A)
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  • General equation of a hyperbola is $\large\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$$=1$
Answer : $\large\frac{x^2}{25}-\frac{y^2}{24}\normalsize =1$
Given data is vertices are $(\pm 5,0)$ and foci $(\pm 7,0)$
The distance between the two vertices is $2a$.
(ie) $V_1V_2=10$
$\Rightarrow 2a=10$
$\therefore a =5$
The distance between the two foci is $2ae$
(ie) $2ae=14$
$ae=7$
$\therefore 5\times e=7$
$\Rightarrow e=\large\frac{7}{5}$
$\therefore b^2=a^2(e^2-1)$
$b^2=25(\large\frac{49}{25}$$-1)$
$\;\;\;\;=25(\large\frac{24}{25})$
$\Rightarrow b^2=24$
Hence the equation of the hyperbola is $\large\frac{x^2}{25}-\frac{y^2}{24}$$=1$.
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