alleryWhat is the eccentricity of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$?
A$\dfrac{4}{3}$
B$\dfrac{5}{3}$
C$\dfrac{5}{4}$
D$\dfrac{3}{5}$
Answer & Solution
Correct answer: B. $\dfrac{5}{3}$
$a^2 = 9, b^2 = 16$, so $a = 3$.
$c^2 = a^2 + b^2 = 25 \Rightarrow c = 5$.
$e = \dfrac{c}{a} = \dfrac{5}{3}$.
Sanity: $e > 1$ — consistent with a hyperbola.
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