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.
Keep practising Mensuration: Surface Areas & Volumes
Browse all Mensuration: Surface Areas & Volumes questions for GRE, or explore more Mathematics topics.
Related questions
A line passes through the points $(1, 2, 3)$ and $(4, 6, 9)$. The direction ratios of the The direction cosines of a line are $l$, $m$, $n$. They satisfy:The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in three-dimensiA fair coin is tossed twice. How many outcomes are there in the sample space?A hyperbola has foci at $(\pm 5, 0)$ and the length of the transverse axis is $6$. Find itWhat is the eccentricity of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$?For any hyperbola, the eccentricity $e$ satisfies:The standard equation of a hyperbola with centre at the origin and foci on the $x$-axis is