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A hyperbola has foci at $(\pm 5, 0)$ and the length of the transverse axis is $6$. Find its equation.
A$\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$
B$\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$
C$\dfrac{x^2}{25} - \dfrac{y^2}{9} = 1$
D$\dfrac{x^2}{9} - \dfrac{y^2}{25} = 1$
Answer & Solution
Correct answer: A. $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$
**Read the geometry.**
- Foci on $x$-axis at $(\pm 5, 0) \Rightarrow c = 5$.
- Transverse axis length $= 2a = 6 \Rightarrow a = 3$, so $a^2 = 9$.
**Recover $b$.** $b^2 = c^2 - a^2 = 25 - 9 = 16$.
**Equation.** $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.
**Sanity check.** Plug in vertex $(3, 0)$: $\tfrac{9}{9} - 0 = 1$ ✓. Eccentricity $e = 5/3 > 1$ ✓.
**Distractor logic.** Option C ($a^2 = 25$) treats the *focal* distance as the semi-transverse axis — but $c \neq a$ for any hyperbola; in fact $c > a$ always.
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