1. **State the problem:** We need to find the value of $a$ such that the line $3y = ax + 9$ is tangent to the curve $y^2 = 49$. 2. **Rewrite the line equation:** Divide both sides by 3 to express $y$ in terms of $x$: $$y = \frac{ax + 9}{3}$$ 3. **Substitute $y$ into the curve equation:** The curve is $y^2 = 49$, so substitute $y$: $$\left(\frac{ax + 9}{3}\right)^2 = 49$$ 4. **Simplify the equation:** $$\frac{(ax + 9)^2}{9} = 49$$ Multiply both sides by 9: $$ (ax + 9)^2 = 441 $$ 5. **Expand the left side:** $$ (ax)^2 + 2 \cdot ax \cdot 9 + 9^2 = 441 $$ $$ a^2 x^2 + 18 a x + 81 = 441 $$ 6. **Bring all terms to one side:** $$ a^2 x^2 + 18 a x + 81 - 441 = 0 $$ $$ a^2 x^2 + 18 a x - 360 = 0 $$ 7. **Condition for tangency:** The line touches the curve, so the quadratic in $x$ has exactly one solution. The discriminant $\Delta$ must be zero: $$ \Delta = (18 a)^2 - 4 \cdot a^2 \cdot (-360) = 0 $$ 8. **Calculate the discriminant:** $$ 324 a^2 + 1440 a^2 = 0 $$ $$ 1764 a^2 = 0 $$ 9. **Solve for $a$:** $$ a^2 = 0 \implies a = 0 $$ 10. **Conclusion:** The value of $a$ for which the line is tangent to the curve is $\boxed{0}$. This means the line is $3y = 9$ or $y = 3$, which is a horizontal line tangent to the circle $y^2 = 49$ at $y = 3$ or $y = -3$.