Question

Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
(a) 16x² − 9y² = 144
(b) 9x² − 16y² = 144
(c) 25x² − 9y² = 225
(d) 9x² − 25y² = 81

Answer 1

(a) 16x² − 9y² = 144

The vertices of the hyperbola are $\left(\pm 3,0\right)$ and foci are $\left(\pm 5,0\right)$.
Thus, the values of a and ae are 3 and 5, respectively.

Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} b^2=25-9 \\ \Rightarrow b^2=16 \end{gathered}$$
Equation of the hyperbola is given below:
$$\begin{gathered} \frac{x^2}{9}-\frac{y^2}{16}=1 \\ \Rightarrow 16x^2-9y^2=144 \end{gathered}$$