Home › JEE Advanced › Mathematics › Conic Sections ›  Find the foci of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.
A$(\pm 3, 0)$
B$(\pm 4, 0)$
C$(\pm 5, 0)$
D$(0, \pm 5)$
Answer & Solution
Correct answer: C. $(\pm 5, 0)$
Identify $a^2 = 9, b^2 = 16$ (transverse axis along $x$). For a hyperbola, $c^2 = a^2 + b^2$ (note the **plus**, unlike ellipse's minus).
$c^2 = 9 + 16 = 25 \Rightarrow c = 5$.
Foci are on the transverse axis at $(\pm c, 0) = (\pm 5, 0)$.
**Trap.** Option A picks $a$, option B picks $b$ — both are axis-endpoint distances, not focal distances. The relation $c^2 = a^2 + b^2$ is the load-bearing identity here.
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