1. **State the problem:** Find the first-quadrant point on the curve defined by the equation $y^2 x = 18$ that is closest to the point $(2,0)$. 2. **Understand the problem:** We want to minimize the distance between a point $(x,y)$ on the curve and the point $(2,0)$. 3. **Distance formula:** The distance squared between $(x,y)$ and $(2,0)$ is $$D^2 = (x-2)^2 + (y-0)^2 = (x-2)^2 + y^2.$$ Minimizing $D^2$ also minimizes $D$, so we minimize $D^2$ for simplicity. 4. **Constraint:** The point $(x,y)$ lies on the curve $y^2 x = 18$, so $$y^2 = \frac{18}{x}.$$ Since we want the first quadrant, $x > 0$ and $y > 0$. 5. **Substitute $y^2$ into $D^2$:** $$D^2 = (x-2)^2 + y^2 = (x-2)^2 + \frac{18}{x}.$$ 6. **Minimize $D^2$ with respect to $x$:** Take derivative: $$\frac{d}{dx} D^2 = 2(x-2) - \frac{18}{x^2} = 0.$$ 7. **Solve for $x$:** $$2(x-2) = \frac{18}{x^2}$$ $$2x - 4 = \frac{18}{x^2}$$ Multiply both sides by $x^2$: $$2x^3 - 4x^2 = 18$$ $$2x^3 - 4x^2 - 18 = 0$$ Divide by 2: $$x^3 - 2x^2 - 9 = 0.$$ 8. **Find roots:** Try $x=3$: $$3^3 - 2(3)^2 - 9 = 27 - 18 - 9 = 0,$$ so $x=3$ is a root. 9. **Find $y$:** $$y^2 = \frac{18}{3} = 6 \implies y = \sqrt{6}$$ (positive since first quadrant). 10. **Answer:** The point is $(3, \sqrt{6})$. **Final answer:** (3, \sqrt{6}) which corresponds to option (C).