[THOUGHT] I will model the hyperbola and tangent line using sympy, compute the intersection point with the y-axis, then compute distances from that point to the two foci, form their product, set it equal to 8, and solve for (a). Then I'll extract (a^2 = p/q) and compute (p+q). [/THOUGHT] [PYTHON] import sympy as sp

Define symbol

a = sp.symbols('a', positive=True)

Define point on hyperbola

x0 = a y0 = sp.sqrt(2)*a

Slope of tangent using implicit differentiation: y^2/a^2 - x^2/a^2 = 1

dy/dx = x / y

m = x0 / y0 m_simplified = sp.simplify(m)

Equation of tangent line: y - y0 = m (x - x0)

x, y = sp.symbols('x y') y_line = y0 + m_simplified*(x - x0)

Intersection with y-axis: x = 0

yP = sp.simplify(y_line.subs(x, 0)) P = (0, yP)

Foci of hyperbola y^2/a^2 - x^2/a^2 = 1: (0, ±a*sqrt(2))

F1 = (0, asp.sqrt(2)) F2 = (0, -asp.sqrt(2))

Distances PF1 and PF2

PF1 = sp.simplify(sp.sqrt((P[0]-F1[0])**2 + (P[1]-F1[1])**2)) PF2 = sp.simplify(sp.sqrt((P[0]-F2[0])**2 + (P[1]-F2[1])**2))

product = sp.simplify(PF1 * PF2) print("Slope m:", m_simplified) print("Point P:", P) print("PF1:", PF1) print("PF2:", PF2) print("Product PF1*PF2:", product)

Set product = 8 and solve for a^2

solution_for_a2 = sp.solve(sp.Eq(product, 8), a**2) print("Solutions for a^2:", solution_for_a2)

Extract p and q from a^2 = p/q and compute p+q

a2 = sp.simplify(solution_for_a2[0]) p = sp.fraction(a2)[0] q = sp.fraction(a2)[1] print("a^2:", a2) print("p:", p, "q:", q, "p+q:", p+q) [/PYTHON]

[THOUGHT] From the Python output, we have (a^2 = \frac{16}{3}), so (p=16) and (q=3), giving (p+q=19). [ \boxed{19} ]