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Find the equation of the hyperbola satisfying the give conditions : Vertices $(\pm 7,\,0)$, $e=\frac{4}{3}$
$\frac{x^{2}}{49}-\frac{y^{2}}{343}=1$
$\frac{x^{2}}{49}-\frac{y^{2}}{343}=1$
$\frac{x^{2}}{49}-\frac{y^{2}}{343}=1$
$\frac{x^{2}}{49}-\frac{y^{2}}{343}=1$
Solution
Vertices $(\pm 7,\,0)$, $e=\frac{4}{3}$
Here, the vertices are on the $x-$ axis.
Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
since the vertices are $(\pm 7,\,0),\,\,a =7$
It is given that $e=\frac{4}{3}$
$\therefore \frac{c}{a}=\frac{4}{3} \,\,\,\left[e=\frac{c}{a}\right]$
$\Rightarrow \frac{c}{7}=\frac{4}{3}$
$\Rightarrow c=\frac{28}{3}$
We know that $a^{2}+b^{2}=c^{2}$
$\therefore 7^{2}+b^{2}=\left(\frac{28}{3}\right)^{2}$
$\Rightarrow b^{2}=\frac{784}{9}-49$
$\Rightarrow b^{2}=\frac{784-441}{9}=\frac{343}{9}$
Thus, the equation of the hyperbola is $\frac{x^{2}}{49}-\frac{y^{2}}{343}=1$
