[THOUGHT] We are given points A(5,0) and B(-5,0). Point C satisfies AC = 6, and P lies on segment AC. Also BP + CP = 8. Let θ = ∠CAB. We place coordinate system as given, set C using polar coordinates from A: C = (5 - 6 cosθ, 6 sinθ). This ensures AC = 6 and ∠CAB = θ.

Parameterize P on AC by AP = x, so P = (5 - x cosθ, x sinθ). Condition BP + CP = 8 yields BP = x+2. Computing BP² from coordinates gives x = 24/(5 cosθ + 1). (Derived from distance equation.)

Area of triangle BCP computed via determinant yields f(θ) = 5 sinθ (6 - x) = 30 sinθ (5 cosθ - 3)/(5 cosθ + 1).

Now differentiate: f'(θ) = 30 [ cosθ·(5C-3)/(5C+1) - 20 sin²θ/(5C+1)² ], where C = cosθ, S = sinθ.

Given β with cosβ = 4/5, sinβ = 3/5. Substituting C = 4/5, S = 3/5 gives

f'(β) = 30[ (4/5·1/5) - 20·(9/25)/25? Wait compute stepwise:

C·(5C-3)/(5C+1) = (4/5·1)/5 = 4/25. 20 S²/(5C+1)² = 20·(9/25)/25 = 180/625 = 36/125? Actually 20·9/(25·25)=180/625=36/125. But careful: we need the term exactly as in f' expression: f' = 30[ C(5C-3)/(5C+1) - 20 S²/(5C+1)² ].

Compute first term: 30·(4/25)=120/25=24/5. Second term: 600·(9/25)/25? Actually 30·20 S²/(5C+1)² = 600·(9/25)/25? Let's use the combined form: f' = 30·C(5C-3)/(5C+1) - 600·S²/(5C+1)².

At C=4/5, S=3/5: 30·C(5C-3)/(5C+1)=30·(4/25)=120/25=24/5. 600·S²/(5C+1)² = 600·(9/25)/25 = 600·9/(625)=5400/625=216/25. Then f' = algorithmically? Actually 30·C(5C-3)/(5C+1) = 24/5, and 600·S²/(5C+1)² = 216/25? Wait 600·9/625 = 5400/625 = 8.64 = 216/25? 216/25 = 8.64, yes. So f' = 24/5 - 216/25 = (120/25 - 216/25) = -96/25. So f'(β) = -96/25.

Thus -p/q = -96/25 → p = 96, q = 2 = 25, which are coprime. Then p + q² = 96 + 625 = 721.

Now we verify with Python to be safe.

[PYTHON] import sympy as sp

Define variable and functions

theta = sp.symbols('theta') C = sp.cos(theta) S = sp.sin(theta)

f(theta) = 30 * S * (5C - 3) / (5C + 1)

f = 30 * S * (5C - 3) / (5C + 1)

Derivative

fprime = sp.diff(f, theta)

Evaluate at theta where cos = 4/5, sin = 3/5

fprime_beta = fprime.subs({sp.cos(theta): sp.Rational(4,5), sp.sin(theta): sp.Rational(3,5)}) fprime_simplified = sp.simplify(fprime_beta) print("f'(β) =", fprime_simplified) # Should be -96/25

Extract p and q

Since f'(β) = -p/q => p/q = -f'(β)

p_over_q = -fprime_simplified p_val = sp.nsimplify(p_over_q) print("p/q =", p_val) # Should be 96/25 p = p_val.p q = p_val.q result = p + q**2 print("p + q^2 =", result) # Should be 721 [/PYTHON]

The output is: f'(β) = -96/25 p/q = 96/25 p + q^2 = 721

Thus the final answer is \boxed{721}.

The problem defines points (A(5,0)) and (B(-5,0)), with point (C) such that (AC = 6) and (P) lies on segment (AC) with (BP + CP = 8). Let (\theta = \angle CAB). The area of (\triangle BCP) is denoted (f(\theta)). For (\beta) with (\cos\beta = \frac{4}{5}) and (0 < \beta < \frac{\pi}{2}), we find (f'(\beta) = -\frac{p}{q}) in lowest terms and need (p + q^2).

Using coordinates, set (C = (5 - 6\cos\theta, 6\sin\theta)) to ensure (AC = 6) and (\angle CAB = \theta). Parameterize (P) on (AC) by (AP = t), so (P = (5 - t\cos\theta, t\sin\theta)). The condition (BP + CP = 8) leads to (t = \frac{24}{1 + 5\cos\theta}). The area of (\triangle BCP) is

[ f(\theta) = \frac{1}{2} \left| \det(\overrightarrow{BC}, \overrightarrow{BP}) \right| = \frac{30\sin\theta ,(5\cos\theta - 3)}{5\cos\theta + 1}. ]

Differentiating (f(\theta)) and evaluating at (\beta) with (\cos\beta = \frac{4}{5}), (\sin\beta = \frac{3}{5}) yields

[ f'(\beta) = -\frac{96}{25}. ]

Thus (p = 96), (q = 25), and

[ p + q^2 = ow.

\boxed{721}