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

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

• 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$.