Find the value of k...derivatives?

Suppose the point of tangency is (p, f(p)) = (p, p² - kp)

From y = 6x - 25, we see that f'(p) = 6 so

2p - k = 6

which implies

k = 2p - 6

The point (p, p² - kp) lies on the line, so it satisfies the line equation:

p² - kp = 6p - 25

p² - (k+6)p + 25 = 0

Substituting for k from the above result,

p² - (2p)p + 25 = 0

-p² + 25 = 0

p = ±5

k = 2p - 6, so if p = 5, then k = 4; if p = -5, then k = -16

f'(x) = 2x - k, so the tangent line at (a, a^2 - ka) has slope 2a -k and equation

y - (a^2 - ka) = (2a - k) (x-a) or

y = (2a - k) x - a(2a - k) + (a^2-ka), which simplifies to

y = (2a-k) x - a^2

To make the given tangent line fit this we need

2a - k = 6 and a^2 = 25, hence a =5 and k = 4, or a= -5 and k = -16