1. **State the problem:** We need to find the value of $a$ such that the line $3y = ax + 9$ is tangent to the curve $y^2 = 4x$. 2. **Rewrite the line equation:** Express $y$ in terms of $x$: $$3y = ax + 9 \implies y = \frac{ax + 9}{3}$$ 3. **Substitute $y$ into the curve equation:** The curve is $y^2 = 4x$. Substitute $y$: $$\left(\frac{ax + 9}{3}\right)^2 = 4x$$ 4. **Simplify the equation:** $$\frac{(ax + 9)^2}{9} = 4x \implies (ax + 9)^2 = 36x$$ 5. **Expand the left side:** $$a^2x^2 + 2 \cdot a \cdot 9 x + 81 = 36x$$ $$a^2 x^2 + 18 a x + 81 = 36 x$$ 6. **Bring all terms to one side:** $$a^2 x^2 + 18 a x + 81 - 36 x = 0$$ $$a^2 x^2 + (18 a - 36) x + 81 = 0$$ 7. **Condition for tangency:** The line touches the curve, so the quadratic in $x$ has exactly one solution. The discriminant $\Delta$ must be zero: $$\Delta = b^2 - 4ac = 0$$ Here, $a = a^2$, $b = 18 a - 36$, $c = 81$. 8. **Calculate the discriminant:** $$\Delta = (18 a - 36)^2 - 4 \cdot a^2 \cdot 81 = 0$$ 9. **Simplify:** $$ (18 a - 36)^2 = 324 a^2$$ 10. **Take square root:** $$|18 a - 36| = 18 |a|$$ 11. **Solve the two cases:** - Case 1: $18 a - 36 = 18 a$ gives $-36 = 0$, no solution. - Case 2: $18 a - 36 = -18 a$ gives $$18 a - 36 = -18 a \implies 36 a = 36 \implies a = 1$$ **Final answer:** $$\boxed{1}$$