[THOUGHT] To solve the problem, we start by parameterizing point ( C ) in terms of the angle ( \theta ). Given ( A(5,0) ) and ( B(-5,0) ), the coordinates of ( C ) are: [ C = (5 - 6 \cos \theta, 6 \sin \theta) ] Point ( P ) lies on segment ( AC ) and can be parameterized as: [ P = (5 - 6t \cos \theta, 6t \sin \theta) ] where ( t ) is a parameter between 0 and 1. Using the condition ( BP + CP = 8 ), we derive the value of ( t ): [ t = \frac{4}{5 \cos \theta + 1} ] Substituting ( t ) into the coordinates of ( P ), we get: [ P = \left(5 - \frac{24 \cos \theta}{5 \cos \theta + 1}, \frac{24 \sin \theta}{5 \cos \theta + 1}\right) ] The area of triangle ( BCP ) is given by: [ f(\theta) = \frac{1}{2} \left| (C_x - B_x)(P_y - B_y) - (P_x - B_x)(C_y - B_y) \right| ] Substituting the coordinates of ( B ), ( C ), and ( P ), we simplify the area expression to: [ f(\theta) = 30 \sin \theta \frac{5 \cos \theta - 3}{5 \cos \theta + 1} ] Next, we compute the derivative ( f'(\theta) ) using the quotient rule and product rule. After simplification, we find: [ f'(\theta) = 30 \frac{25 \cos^3 \theta + 10 \cos^2 \theta - 3 \cos \theta - 20}{(5 \cos \theta + 1)^2} ] Evaluating ( f'(\theta) ) at ( \theta = \beta ) where ( \cos \beta = \frac{4}{5} ), we substitute ( \cos \beta = \frac{4}{5} ) and ( \sin \beta = \frac{3}{5} ) into the derivative: [ f'(\beta) = 30 \frac{25 \left(\frac{4}{5}\right)^3 + 10 \left(\frac{4}{5}\right)^2 - 3 \left(\frac{4}{5}\right) - 20}{\left(5 \left(\frac{4}{5}\right) + 1\right)^2} ] Simplifying the numerator and denominator, we get: [ f'(\beta) = 30 \frac{25 \cdot \frac{64}{125} + 10 \cdot \frac{16}{25} - 3 \cdot \frac{4}{5} - 20}{(4 + 1)^2} = 30 \frac{12.8 + 6.4 - 2.4 - 20}{25} = 30 \frac{-3.2}{25} = -\frac{96}{25} ] Thus, ( p = 96 ) and ( q = 25 ). Therefore, ( p + q^2 = 96 + 25^2 = 96 + 625 = 721 ).

[PYTHON]

Verification of the derivative calculation

import sympy as sp

theta = sp.symbols('theta') cos_beta = sp.Rational(4, 5) sin_beta = sp.sqrt(1 - cos_beta**2)

Define the function f(theta)

f = 30 * sp.sin(theta) * (5 * sp.cos(theta) - 3) / (5 * sp.cos(theta) + 1)

Compute the derivative f'(theta)

f_prime = sp.diff(f, theta)

Substitute theta = beta where cos(beta) = 4/5

f_prime_beta = f_prime.subs({sp.cos(theta): cos_beta, sp.sin(theta): sin_beta})

Simplify the result

f_prime_beta_simplified = sp.simplify(f_prime_beta) print(f_prime_beta_simplified) [/PYTHON]

[THOUGHT] The Python code confirms that the derivative ( f'(\beta) = -\frac{96}{25} ). Therefore, ( p = 96 ) and ( q = 25 ), leading to ( p + q^2 = 96 + 625 = 721 ).

[ \boxed{721} ]

[THOUGHT] The Python output confirms ( f'(\beta) = -\frac{96}{25} ), so ( p = 96 ) and ( q = 25 ). Thus, ( p + q^2 = 96 + 25^2 = 96 + 625 = 721 ).

[ \boxed{721} ]