Question

Find the equation of the hyperbola whose vertices are $(\pm 7,0)$ and the eccentricity is $\frac{4}{3}.$

Answer 1

Since the vertices of the given hyperbola are of the form $(\pm a,0),$ it is a horizontal hyperbola.

Let the required equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

Then, its vertices are $(\pm a,0).$

But, the vertices are ($\pm 7,0).$

$\therefore a=7\iff a^2=49.$

Also, $e=\frac{c}{a}\iff c=ae=(7\times \frac{4}{3})=\frac{28}{3}.$

Now, $c^2=(a^2+b^2)\iff b^2-(c^2-a^2)=[{\left(\frac{28}{3}\right)}^2-49]=\frac{343}{9}.$

Thus, $a^2=49$ and $b^2=\frac{343}{9}.$

$\therefore$ the required equation is,

$\frac{x^2}{49}-\frac{y^2}{\left(343/9\right)}=1\iff \frac{x^2}{49}-\frac{9y^2}{343}=1\iff 7x^2-9y^2=343.$