[THOUGHT] Compute tangent intersection point P, then distances from P to the foci and enforce PF*PF' = 8 to solve for a^2 and p+q. Verify with Python. [/THOUGHT] [PYTHON] import sympy as sp

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

foci for y^2 - x^2 = a^2: (0, ±a*sqrt(2))

F = sp.Point(0, asp.sqrt(2)) Fp = sp.Point(0, -asp.sqrt(2))

tangent at (a, sqrt(2)a) for y^2 - x^2 = a^2: slope x/y = 1/sqrt(2)

P_tan = sp.Point(a, a*sp.sqrt(2)) m = sp.Rational(1,1)/sp.sqrt(2)

y-intercept at x=0:

yP = P_tan.y + m*(0 - P_tan.x) P = sp.Point(0, sp.simplify(yP))

PF = P.distance(F) PFp = P.distance(Fp) prod = sp.simplify(PF*PFp)

print("P =", P) print("PF =", sp.simplify(PF)) print("PF' =", sp.simplify(PFp)) print("Product =", prod)

sol_a2 = sp.solve(sp.Eq(prod, 8), a**2)[0] print("a^2 =", sp.simplify(sol_a2)) p,q = sp.fraction(sp.simplify(sol_a2)) print("p+q =", sp.simplify(p+q)) [/PYTHON][THOUGHT] Use the verified computation to extract $a^2$ and then $p+q$.

$\boxed{19}$

[THOUGHT] From the Python output, $PF\cdot PF'=\frac{3a^2}{2}=8$, so $a^2=\frac{16}{3}$. Thus $p+q=16+3=19$. $\boxed{19}$