The curve C has equation $y = \frac{1}{3} x^2 + 8$\n\nThe line L has equation $y = 3x + k$, where $k$ is a positive constant.\n\n(a) Sketch C and L on separate diagrams, showing the coordinates of the points at which C and L cut the axes.\n\nGiven that line L is a tangent to C,\n(b) find the value of k. - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

We know line L is tangent to curve C, meaning they intersect at exactly one point.

1. Equate the functions:
   $\frac{1}{3} x^2 + 8 = 3x + k$

2. Rearranging gives:
   $\frac{1}{3} x^2 - 3x + (8 - k) = 0$

3. For tangency, the discriminant must be zero:
   $b^2 - 4ac = 0$
   Here, $a = \frac{1}{3}$, $b = -3$, $c = 8 - k$. Thus:
   $(-3)^2 - 4 \cdot \frac{1}{3} \cdot (8 - k) = 0$

4. Simplifying gives:
   $9 - \frac{4}{3} (8 - k) = 0 \Rightarrow 9 = \frac{32}{3} - \frac{4k}{3}$

5. Multiplying through by 3 to eliminate fractions:
   $27 = 32 - 4k \Rightarrow 4k = 32 - 27 \Rightarrow 4k = 5 \Rightarrow k = \frac{5}{4}$

Thus, the value of k is ( \frac{5}{4} ).