# Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0),

Question:

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

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 $\frac{x^{2}}{49}-\frac{9 y^{2}}{343}=1$.