A parabola has its vertex at the origin, its axis along the positive $x$-axis, and passes through the point $(4, 6)$. Find its equation.

**Setup.** Axis along positive $x$-axis with vertex at origin ⇒ standard form $y^2 = 4ax$. **Plug in the known point.** $(4, 6)$ on the curve: $6^2 = 4a \cdot 4 \Rightarrow 36 = 16a \Rightarrow a = \dfrac{9}{4}$. **Substitute back.** $y^2 = 4 \cdot \dfrac{9}{4} \cdot x = 9x$. Quick verification: $(4, 6)$ ⇒ $36 = 9 \times 4 = 36$ ✓. **Distractor check.** Option B ($y^2 = 4x$) corresponds to $a = 1$ — would pass through $(4, 4)$, not $(4, 6)$. Option D ($a = 9$) overshoots: $(4, ?) \Rightarrow y^2 = 144$, $y = 12$.