Question

Find the equation of the hyperbola satisfying the give conditions: Vertices $(\pm 7,0)$
$e=\frac{4}{3}$

Answer 1

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 $(\pm 7,0)\Rightarrow a=7$
It is given that $e=\frac{4}{3}$
$\therefore \frac{c}{a}=\frac{4}{3},$ $[\because e=\frac{c}{a}]$
$\Rightarrow \frac{c}{7}=\frac{4}{3}$
$\Rightarrow c=\frac{28}{3}$
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 $\frac{x^2}{49}-\frac{9y^2}{343}=1$