[THOUGHT] We need to solve for a > 0 given the hyperbola: x^2/a^2 - y^2/a^2 = -1, which can be rewritten as y^2/a^2 - x^2/a^2 = 1. That's a vertical hyperbola centered at origin with semi-major axis a (vertical) and semi-minor axis a (horizontal). Actually standard form: y^2 / a^2 - x^2 / a^2 = 1. So it's a rectangular hyperbola? Wait, both denominators equal a^2, so asymptotes are y = ± x. But foci: for vertical hyperbola, c^2 = a^2 + b^2 = a^2 + a^2 = 2a^2, so c = a√2. Foci are at (0, ±c) = (0, ±a√2). So F and F' are (0, a√2) and (0, -a√2). However the problem says foci are F and F', but doesn't specify which is which. We'll keep them as (0, c) and (0, -c) with c = a√2.

Point on hyperbola: (a, √2 a). Check: y^2/a^2 - x^2/a^2 = (2a^2)/a^2 - a^2/a^2 = 2 - 1 = 1, yes.

Tangent line at (a, √2 a): For hyperbola y^2/a^2 - x^2/a^2 = 1, differentiate implicitly: (2y/a^2) dy - (2x/a^2) dx = 0 => dy/dx = (x)/(y). At (a, √2 a), slope m = a/(√2 a) = 1/√2.

Equation: y - √2 a = (1/√2)(x - a). Intersect with y-axis (x=0): y = √2 a + (1/√2)(-a) = √2 a - a/√2 = (2a/√2 - a/√2) = a/√2. So P = (0, a/√2).

Now PF = distance from P to F. F = (0, a√2). So PF = |a√2 - a/√2| = a|√2 - 1/√2| = a| (2/√2 - 1/√2) | = a| (1/√2) | = a/√2.

Similarly, PF' = distance from P to F' = | -a√2 - a/√2| = a| -√2 - 1/√2| = a| -(√2 + 1/√2)| = a(√2 + 1/√2) = a( (2/√2 + 1/√2) ) = a(3/√2) = 3a/√2.

Thus product PF × PF' = (a/√2) * (3a/√2) = 3a^2 / 2.

Given product = 8, so 3a^2 / 2 = 8 => a^2 = 16/3.

Thus a^2 = 16/3 = p/q, p=16, q=3, p+q = 19.

But let's verify with Python to avoid mistakes.

We'll compute tangent, intersection, distances, product, solve for a.

We'll use sympy to solve.

But first, ensure hyperbola equation is as given: x^2/a^2 - y^2/a^2 = -1 => y^2/a^2 - x^2/a^2 = 1.

Yes.

We'll do symbolic steps.

Let's write Python code to compute everything and solve.

We'll define a as symbol, find c = sqrt(2)*a, point (a, sqrt(2)*a), slope = derivative, tangent line equation, intersection with y-axis, distances PF, PF', product, solve product=8 for a>0.

Then compute a^2, express as fraction p/q, compute p+q.

Let's do it. </think>

[PYTHON] import sympy as sp

Define symbols

a = sp.symbols('a', positive=True, real=True) c = sp.sqrt(2) * a # focal distance

Point on hyperbola

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

Slope of tangent: dy/dx = x/y

slope = x0 / y0 # simplifies to a/(sqrt(2)*a) = 1/sqrt(2)

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

x, y = sp.symbols('x y') tangent_eq = y - y0 - slope*(x - x0)

Intersection with y-axis: set x = 0

y_intercept = sp.solve(tangent_eq.subs(x, 0), y)[0]

Should be a/√2

print("P = (0, y_intercept):", (0, y_intercept)) print("Simplified y_intercept:", sp.simplify(y_intercept))

Distances PF and PF'

F1 = (0, c) F2 = (0, -c) PF = sp.sqrt((0 - 0)**2 + (y_intercept - c)**2) PF_prime = sp.sqrt((0 - 0)**2 + (y_intercept + c)**2) # since -c to y_intercept: y_intercept - (-c) = y_intercept + c

print("PF =", PF) print("PF' =", PF_prime) print("PF * PF' =", sp.simplify(PF * PF_prime))

Set product equal to 8 and solve for a

equation = sp.Eq(PF * PF_prime, 8) solutions = sp.solve(equation, a) print("Solutions for a:", solutions)

a > 0, take positive real solution

a_solution = [sol for sol in solutions if sol > 0][0] print("a =", a_solution) print("a^2 =", a_solution**2)

Express a^2 as rational p/q

a2_expr = sp.simplify(a_solution**2)

Convert to rational if possible

p, q = sp.fraction(a2_expr) print("p =", p, "q =", q) print("p + q =", p + q) [/PYTHON]

[THOUGHT] The Python output gives a^2 = 16/3, so p = 16, q = 3, and p + q = 19.

\boxed{19}