[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:

Point $P$ lies on segment $AC$ and can be parameterized as:

where $t$ is a parameter between 0 and 1. Using the condition $BP + CP = 8$, we derive the value of $t$:

Substituting $t$ into the coordinates of $P$, we get:

The area of triangle $BCP$ is given by:

Substituting the coordinates of $B$, $C$, and $P$, we simplify the area expression to:

Next, we compute the derivative $f'(\theta)$ using the quotient rule and product rule. After simplification, we find:

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:

Simplifying the numerator and denominator, we get:

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 $p + q^2 = 96 + 625 = 721$.

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