Question

The foci of the hyperbola$9x^2-16y^2=144$ are

• A. $(\pm 4,0)$
• B. $(0,\pm 4)$
• C. $(\pm 5,0)$
• D. $(0,\pm 5)$

Answer 1

The correct option is C

$(\pm 5,0)$

$\text{The equation of the hyperbola is given below}$

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

$\text{This equation can be rewritten in the following way:}$

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

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

$Thisisthestandardequationofahyperbola,where{a}^2=16and{b}^2=9$

$\text{This eccintricity is calculated in the following way :}$

$b^2=a^2(e^2-1)$

$\Rightarrow 9=16(e^2-1)$

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

$\Rightarrow e=\frac{5}{4}$

$Foci=(\pm ae,0)=(\pm 5,0)$