[THOUGHT] We are given the hyperbola equation: ( \frac{x^2}{a^2} - \frac{y^2}{a^2} = -1 ). Multiply by (a^2): (x^2 - y^2 = -a^2), or equivalently (y^2 - x^2 = a^2). This is a hyperbola centered at the origin with transverse axis along the y-axis (since positive term on y^2). Standard form: (\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1). Indeed, dividing by (a^2) gives (\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1). So it's a rectangular hyperbola? Actually (a^2) same for both axes, so asymptotes slopes ±1.

Foci: For hyperbola (\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1), distance from center to foci along y-axis: (c = \sqrt{a^2 + a^2} = a\sqrt{2}). So foci are at ((0, \pm a\sqrt{2})). Let (F = (0, a\sqrt{2})) and (F' = (0, -a\sqrt{2})).

Point on hyperbola: ((a, \sqrt{2}a)). Check: plug into equation: (a^2/a^2 - (\sqrt{2}a)^2/a^2 = 1 - 2 = -1), yes satisfies.

Tangent line at that point: We can find derivative implicitly. Given (x^2 - y^2 = -a^2). Differentiate: (2x - 2y y' = 0) → (y' = x/y). At ((a, \sqrt{2}a)), slope (m = a/(\sqrt{2}a) = 1/\sqrt{2}). So tangent line: (y - \sqrt{2}a = (1/\sqrt{2})(x - a)). Find intersection with y-axis (x=0): (y = \sqrt{2}a + (1/\sqrt{2})(-a) = \sqrt{2}a - a/\sqrt{2} = \sqrt{2}a - (\sqrt{2}/2)a?) Wait: (a/\sqrt{2} = a\sqrt{2}/2). So (y = \sqrt{2}a - \frac{\sqrt{2}}{2}a = \frac{\sqrt{2}}{2}a(2 - 1) = \frac{\sqrt{2}}{2}a = a/\sqrt{2}). So P = (0, a/√2).

Now distances PF and PF': F = (0, a√2), F' = (0, -a√2). PF = distance between (0, a/√2) and (0, a√2): vertical distance = |a√2 - a/√2| = a|√2 - 1/√2| = a| (2 - 1)/√2 | = a(1/√2) = a/√2. PF' = distance between (0, a/√2) and (0, -a√2): vertical distance = |a/√2 - (-a√2)| = a/√2 + a√2 = a(1/√2 + √2) = a(1/√2 + √2) = a( (1 + 2)/√2 ) = a(3/√2) = 3a/√2.

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

Given PF × PF' = 8, so 3a^2 / 2 = 8 → a^2 = 16/3 → a = 4/√3 (positive real). So a^2 = 16/3.

Then problem says: If (a^2 = \dfrac{p}{q}), compute (p+q). So p=16, q=3, p+q=19.

But wait: check if P is indeed (0, a/√2). Let's compute numerically with Python to verify distances and product.

Also need to ensure we didn't miss sign: PF = |a√2 - a/√2| = a(√2 - 1/√2) = a( (2-1)/√2 ) = a/√2. Yes positive. PF' = |a/√2 - (-a√2)| = a/√2 + a√2 = a(1/√2 + √2) = a( (1+2)/√2 ) = 3a/√2. So product = 3a^2/2.

Set equal to 8: a^2 = 16/3.

Let's verify with Python. [/THOUGHT] [PYTHON] import sympy as sp

define a as positive real

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

point on hyperbola: (a, sqrt(2)*a)

tangent slope: derivative implicitly: x^2 - y^2 = -a^2

differentiate: 2x - 2y*y' = 0 => y' = x/y

slope = a / (sp.sqrt(2)*a) # simplifies to 1/sqrt(2) slope_simplified = sp.simplify(slope) print("slope:", slope_simplified)

tangent line: y - sqrt(2)a = slope(x - a)

intersection with y-axis: x=0

y_intercept = sp.sqrt(2)*a + slope * (-a) y_intercept_simplified = sp.simplify(y_intercept) print("y-coordinate of P:", y_intercept_simplified)

foci: (0, asqrt(2)) and (0, -asqrt(2))

F = (0, asp.sqrt(2)) Fprime = (0, -asp.sqrt(2)) P = (0, y_intercept_simplified)

distances PF and PF'

PF = sp.sqrt((P[0] - F[0])**2 + (P[1] - F[1])**2) PFprime = sp.sqrt((P[0] - Fprime[0])**2 + (P[1] - Fprime[1])**2) PF_simplified = sp.simplify(PF) PFprime_simplified = sp.simplify(PFprime) print("PF:", PF_simplified) print("PF':", PFprime_simplified)

product

product = sp.simplify(PF_simplified * PFprime_simplified) print("PF * PF':", product)

set equal to 8 and solve for a^2

equation = sp.Eq(product, 8) solution = sp.solve(equation, a**2) print("a^2 =", solution)

solution should be 16/3

print("a^2 simplified:", [sp.nsimplify(sol) for sol in solution])

compute p+q if a^2 = p/q

a_sq = solution[0] p = sp.numer(a_sq) q = sp.denom(a_sq) print("p =", p, "q =", q) print("p+q =", p + q) [/PYTHON]

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