Question

The eccentricity of the conic 9x² − 16y² = 144 is
(a) $\frac{5}{4}$

(b) $\frac{4}{3}$

(c) $\frac{4}{5}$

(d) $\sqrt{7}$

Answer 1

(a) $\frac{5}{4}$

Standard form of a hyperbola = $\frac{x^2}{16}-\frac{y^2}{9}=1$
Here, $a^2=16\mathrm{and}{y}^2=9$

The eccentricity is calculated in the following way:
$$\begin{gathered} b^2=a^2\left(e^2-1\right) \\ \Rightarrow 9=16\left(e^2-1\right) \\ \Rightarrow e^2-1=\frac{9}{16} \\ \Rightarrow e^2=\frac{25}{16} \\ \Rightarrow e=\frac{5}{4} \end{gathered}$$