Question

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

Answer 1

Here vertices are $(\pm 7,0)$ which lie on x - axis.
So the equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore \text{Vertices}(\pm a,0)\text{is}(\pm 7,0)\Rightarrow a=7\text{Now}e=\frac{4}{3}\Rightarrow \frac{c}{a}=\frac{4}{3}\Rightarrow \frac{c}{7}=\frac{4}{3}\Rightarrow c=\frac{28}{3}\text{We know that}{c}^2=a^2+b^2\therefore {\left(\frac{28}{3}\right)}^2=(7{)}^2+b^2\Rightarrow b^2=\frac{784}{9}-49=\frac{343}{9}\text{Thus required equation of hyperbola is},\frac{x^2}{(7{)}^2}-\frac{y^2}{\frac{343}{9}}=1\Rightarrow \frac{x^2}{49}-\frac{y^2}{343}=1.$