alleryThe foci of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ lie at:
A$(\pm 4, 0)$
B$(\pm 3, 0)$
C$(0, \pm 3)$
D$(\pm 5, 0)$
Answer & Solution
Correct answer: B. $(\pm 3, 0)$
For a horizontal ellipse the foci sit at $(\pm c, 0)$ on the major axis, where $c^2 = a^2 - b^2$.
$c^2 = 25 - 16 = 9 \Rightarrow c = 3$. So the foci are at $(\pm 3, 0)$.
Option D mistakenly picks the *vertices*
option A switches $a$ and $b$.
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