[THOUGHT] Let me solve this step by step.

First, I need to find the point P where the curve C intersects the line y = 3x - 5e.

The parametric equations are:

• x = e^(4t)(1 + sin²(πt))
• y = e^(4t)(1 - 3cos²(πt))

For the intersection with y = 3x - 5e: e^(4t)(1 - 3cos²(πt)) = 3·e^(4t)(1 + sin²(πt)) - 5e

Let me simplify and solve for t, then find the slope of the tangent at that point.

The slope of the tangent is dy/dx = (dy/dt)/(dx/dt).

Let me compute this using Python. [/THOUGHT] [PYTHON] import sympy as sp import numpy as np

t = sp.Symbol('t') pi = sp.pi e = sp.E

Parametric equations

x = sp.exp(4t) * (1 + sp.sin(pit)**2) y = sp.exp(4t) * (1 - 3sp.cos(pi*t)**2)

Line equation: y = 3x - 5e

Intersection: e^(4t)(1 - 3cos²(πt)) = 3·e^(4t)(1 + sin²(πt)) - 5e

Let's simplify the intersection equation

e^(4t)(1 - 3cos²(πt)) - 3·e^(4t)(1 + sin²(πt)) + 5e = 0

e^(4t)[(1 - 3cos²(πt)) - 3(1 + sin²(πt))] + 5e = 0

e^(4t)[1 - 3cos²(πt) - 3 - 3sin²(πt)] + 5e = 0

e^(4t)[-2 - 3(cos²(πt) + sin²(πt))] + 5e = 0

e^(4t)[-2 - 3] + 5e = 0

-5·e^(4t) + 5e = 0

e^(4t) = e

4t = 1

t = 1/4

t_val = sp.Rational(1, 4) print(f"t = {t_val}")

Verify the intersection

x_val = x.subs(t, t_val) y_val = y.subs(t, t_val) print(f"x = {x_val}") print(f"y = {y_val}") print(f"x simplified = {sp.simplify(x_val)}") print(f"y simplified = {sp.simplify(y_val)}")

Check if y = 3x - 5e

line_check = 3x_val - 5e print(f"3x - 5e = {sp.simplify(line_check)}") print(f"y = {sp.simplify(y_val)}") print(f"Equal? {sp.simplify(y_val - line_check) == 0}") [/PYTHON]

[THOUGHT] Now I need to find the slope of the tangent at t = 1/4. Let me compute dy/dx = (dy/dt)/(dx/dt) at t = 1/4. [/THOUGHT] [PYTHON] import sympy as sp

t = sp.Symbol('t') pi = sp.pi e = sp.E

Parametric equations

x = sp.exp(4t) * (1 + sp.sin(pit)**2) y = sp.exp(4t) * (1 - 3sp.cos(pi*t)**2)

Derivatives

dx_dt = sp.diff(x, t) dy_dt = sp.diff(y, t)

print(f"dx/dt = {dx_dt}") print(f"dy/dt = {dy_dt}")

Evaluate at t = 1/4

t_val = sp.Rational(1, 4)

dx_dt_val = dx_dt.subs(t, t_val) dy_dt_val = dy_dt.subs(t, t_val)

print(f"\nAt t = 1/4:") print(f"dx/dt = {dx_dt_val}") print(f"dy/dt = {dy_dt_val}") print(f"dx/dt simplified = {sp.simplify(dx_dt_val)}") print(f"dy/dt simplified = {sp.simplify(dy_dt_val)}")

Slope = dy/dx = (dy/dt)/(dx/dt)

slope = dy_dt_val / dx_dt_val slope_simplified = sp.simplify(slope) print(f"\nSlope = {slope_simplified}")

Let's also expand to see the form

slope_expanded = sp.expand(slope_simplified) print(f"Slope expanded = {slope_expanded}")

Try to get it in the form (a*pi - b)/(pi + c)

Let's rationalize or simplify further

slope_rational = sp.together(slope_simplified) print(f"Slope together = {slope_rational}") [/PYTHON]

[THOUGHT] The slope is (3π - 2)/(π + 6), which can be written as (aπ - b)/(π + c) where a = 3, b = 2, and c = 6.

Therefore, a + b + c = 3 + 2 + 6 = 11. [/THOUGHT]

\boxed{11}