Question

$\text{Write the coordinates of the foci of the hyperbola}9x^2-16y^2=144$

Answer 1

$Wehave,$

$9x^2-16y^2=144$

$\Rightarrow \frac{9x^2}{144}-\frac{16y^2}{144}=1$

$\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1$

$\Rightarrow \frac{x^2}{(4{)}^2}-\frac{y^2}{(3{)}^2}=1$

$ItIsoftheform\frac{x^2}{a^2}-\frac{y^2}{b^2}=2,where{a}^2=4and{b}^2=3$

$Now,$

$e=\sqrt{1+\frac{b^2}{a^2}}$

$=\sqrt{1+\frac{9}{16}}$

$=\sqrt{\frac{25}{16}}$

$=\frac{5}{4}$

$Thecoordinatesoffociare(\pm ae,0)i.e.,(\pm 5,0).$