[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 $p+q=19$.