alleryFind the eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.
A$\dfrac{3}{4}$
B$\dfrac{3}{5}$
C$\dfrac{4}{5}$
D$\dfrac{5}{4}$
Answer & Solution
Correct answer: B. $\dfrac{3}{5}$
Identify: $a^2 = 25$ (the larger denominator), $b^2 = 16$. So $a = 5, b = 4$.
$c^2 = a^2 - b^2 = 25 - 16 = 9 \Rightarrow c = 3$.
Eccentricity $e = \dfrac{c}{a} = \dfrac{3}{5}$.
For any ellipse $0 < e < 1$ (a circle is the $e = 0$ limit). Option D ($5/4 > 1$) describes a *hyperbola*, not an ellipse.
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