[THOUGHT] The hyperbola is (\frac{x^2}{a^2} - \frac{y^2}{a^2} = -1), equivalent to (\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1). Point ((a, \sqrt{2}a)) is on it. Slope of tangent: (\frac{dy}{dx} = \frac{x}{y} = \frac{1}{\sqrt{2}}). Equation: (y - \sqrt{2}a = \frac{1}{\sqrt{2}}(x - a)). At (x=0), (y = a / \sqrt{2}), so (P = (0, a / \sqrt{2})).

Foci at ((0, \pm a\sqrt{2})). (PF = |a\sqrt{2} - a/\sqrt{2}| = a(\sqrt{2} - 1/\sqrt{2}) = a / \sqrt{2}).

(PF' = | -a\sqrt{2} - a/\sqrt{2} | = a(\sqrt{2} + 1/\sqrt{2})).

Product: (\frac{a}{\sqrt{2}} \cdot a \left( \sqrt{2} + \frac{1}{\sqrt{2}} \right) = a^2 \left(1 + \frac{1}{2}\right) = \frac{3}{2} a^2 = 8).

Thus, (a^2 = \frac{16}{3}), so (p=16), (q=3), (p+q=19).

Verify the product symbolically with sympy. [/THOUGHT] [PYTHON] import sympy as sp

a = sp.symbols('a', positive=True) PF = a / sp.sqrt(2) PF_prime = a * (sp.sqrt(2) + 1/sp.sqrt(2)) product = PF * PF_prime print(product.simplify())

Solve for a^2

eq = product - 8 sol = sp.solve(eq, a**2) print(sol) print('p+q:', 16 + 3) [/PYTHON]

[THOUGHT] The Python output confirms the product is ( \frac{3}{2} a^2 = 8 ), so ( a^2 = \frac{16}{3} ), and ( p + q = 16 + 3 = 19 ). [/THOUGHT] \boxed{19}

Explanation

The provided AI-generated answer derives from verifying the hyperbola's properties and the given condition through symbolic computation. The hyperbola equation (\frac{x^2}{a^2} - \frac{y^2}{a^2} = -1) rewrites as (\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1), confirming the point ((a, \sqrt{2}a)) lies on it.

Differentiate implicitly to find the tangent slope: (\frac{2x}{a^2} - \frac{2y}{a^2} \frac{dy}{dx} = 0), so (\frac{dy}{dx} = \frac{x}{y}). At ((a, \sqrt{2}a)), the slope is (\frac{a}{\sqrt{2}a} = \frac{1}{\sqrt{2}}).

The tangent line equation is (y - \sqrt{2}a = \frac{1}{\sqrt{2}}(x - a)). At (x=0), substitute to get (y = \sqrt{2}a - \frac{1}{\sqrt{2}}a = a(\sqrt{2} - \frac{1}{\sqrt{2}}) = \frac{a}{\sqrt{2}}), so (P = (0, \frac{a}{\sqrt{2}})).

For this hyperbola, foci are at ((0, \pm c)) where (c = a\sqrt{2}), so (F = (0, a\sqrt{2})) and (F' = (0, -a\sqrt{2})).

Distances: (\overline{PF} = |a\sqrt{2} - \frac{a}{\sqrt{2}}| = a(\sqrt{2} - \frac{1}{\sqrt{2}}) = \frac{a}{\sqrt{2}}), and (\overline{PF'} = |-a\sqrt{2} - \frac{a}{\sqrt{2}}| = a(\sqrt{2} + \frac{1}{\sqrt{2}})).

Product: (\overline{PF} \times \overline{PF'} = \frac{a}{\sqrt{2}} \cdot a \left( \sqrt{2} + \frac{1}{\sqrt{2}} \right) = a^2 \left( \frac{1}{\sqrt{2}} \cdot \sqrt{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \right) = a^2 \left( 1 + \frac{1}{2} \right) = \frac{3}{2} a^2).

Set equal to 8: (\frac{3}{2} a^2 = 8), so (a^2 = \frac{16}{3}). Thus, (a^2 = \frac{p}{q}) where (p=16), (q=3), and (p+q=19).

The Python output verifies this: the product simplifies to (3a^2/2), solving (\frac{3}{2}a^2 - 8 = 0) yields (a^2 = \frac{16}{3}), confirming (p+q=19).