Question

The foci of the hyperbola 9x² − 16y² = 144 are
(a) (± 4, 0)
(b) (0, ± 4)
(c) (± 5, 0)
(d) (0, ± 5)

Answer 1

(c) (± 5, 0)

The equation of the hyperbola is given below:
$9x^2-16y^2=144$
This equation can be rewritten in the following way:
$$\begin{gathered} \frac{9x^2}{144}-\frac{16y^2}{144}=1 \\ \Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1 \end{gathered}$$
This is the standard equation of a hyperbola, where $a^2=16\mathrm{and}{b}^2=9$.

The eccentricity is calculated in the following way:
$$\begin{gathered} b^2=a^2(e^2-1) \\ \Rightarrow 9=16(e^2-1) \\ \Rightarrow \frac{9}{16}=e^2-1 \\ \Rightarrow e=\frac{5}{4} \end{gathered}$$
Foci = $\left(\pm ae,0\right)=\left(\pm 5,0\right)$