The curve C has equation $y = \frac{1}{3}x^2 + 8$ The line L has equation $y = 3x + k$, where k is a positive constant - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

Since L is a tangent to C, we need to find where they touch.

Set the equations equal:

$\frac{1}{3}x^2 + 8 = 3x + k$

Rearranging gives:

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

For this quadratic to have exactly one solution (tangency), the discriminant must be zero:

$b^2 - 4ac = 0$

Here, a = $\frac{1}{3}$, b = -3, and c = $(8 - k)$.

The discriminant is:

$(-3)^2 - 4 \left(\frac{1}{3}\right)(8 - k) = 0$

Simplifying:

$9 - \frac{4}{3}(8 - k) = 0$

Multiply by 3 to eliminate the fraction:

$27 - 4(8 - k) = 0$

Expanding gives:

$27 - 32 + 4k = 0$

$4k = 5$

Thus:

$k = \frac{5}{4}$

Therefore, the value of k is $\frac{5}{4}$.