Question

Write the coordinates of the foci of the hyperbola 9x² − 16y² = 144.

Answer 1

Equation of the hyperbola:
$9x^2-16y^2=144$ ..... (1)

This equation (1) can be rewritten in the following way:
$$\begin{gathered} \Rightarrow \frac{9x^2}{144}-\frac{16y^2}{144}=1 \\ \Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1 \end{gathered}$$

This becomes a standard form of the hyperbola with transverse axis $a=4$ and conjugate axis $b=3$.

$$\begin{gathered} \mathrm{Eccentricity}\mathrm{of}\mathrm{the}hyperbola,e=\frac{\sqrt{a^2+b^2}}{a} \\ =\frac{\sqrt{4^2+3^2}}{4} \\ =\frac{\sqrt{16+9}}{4} \\ =\frac{5}{4} \end{gathered}$$
The foci of the hyperbola are of the form $\left(ae,0\right)$ and $\left(-ae,0\right)$.
Therefore, the foci are $\left(5,0\right)$ and $\left(-5,0\right)$.4