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 $b$.
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