An ellipse has foci at $(\pm 4, 0)$ and passes through the point $(5, 0)$. Find its standard equation.

**Read the geometry.** Foci on the $x$-axis at $(\pm 4, 0)$ ⇒ horizontal ellipse with $c = 4$. The point $(5, 0)$ lies on the major axis ($y = 0$) and on the curve ⇒ it must be a *vertex*, giving $a = 5$. **Recover $b$.** $b^2 = a^2 - c^2 = 25 - 16 = 9$, so $b = 3$. **Equation:** $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$. **Distractor check.** Option B has $b = 4$ — that would put the foci at $(\pm 3, 0)$, not the given $(\pm 4, 0)$.