[THOUGHT] I need to find the value of ( a ) for the given hyperbola. The hyperbola is (\dfrac{x^2}{a^2} - \dfrac{y^2}{a^2} = -1), which can be rewritten as (\dfrac{y^2}{a^2} - \dfrac{x^2}{a^2} = 1). This is a vertical hyperbola centered at the origin. The foci ( F ) and ( F' ) are located at ((0, \pm c)), where ( c = \sqrt{a^2 + a^2} = a\sqrt{2} ).

The point ( P ) is where the tangent to the hyperbola at ((a, \sqrt{2}a)) meets the ( y )-axis. First, I need to find the equation of the tangent line at this point. Then, I can find the coordinates of ( P ). Finally, I need to compute the distances ( \overline{PF} ) and ( \overline{PF'} ), multiply them, and set the product equal to 8 to solve for ( a ).

[/THOUGHT] [PYTHON] from sympy import symbols, Eq, solve, diff, sqrt

a = symbols('a', positive=True) x, y = symbols('x y')

Equation of the hyperbola: y^2/a^2 - x^2/a^2 = 1

hyperbola_eq = Eq(y2/a2 - x2/a2, 1)

Point of tangency: (a, sqrt(2)*a)

point = (a, sqrt(2)*a)

Differentiate implicitly to find the slope of the tangent

dy_dx = diff(hyperbola_eq.lhs, x).subs(hyperbola_eq.rhs, 1) dy_dx = solve(dy_dx, dy_dx[0])[0] # Solve for dy/dx

Slope at the point (a, sqrt(2)*a)

slope = dy_dx.subs({x: point[0], y: point[1]}).simplify()

Equation of the tangent line: y - y1 = m(x - x1)

tangent_line = Eq(y - point[1], slope*(x - point[0]))

Find the y-intercept (P) by setting x = 0

P_y = tangent_line.rhs.subs(x, 0) + point[1] P = (0, P_y)

Foci F and F' are at (0, asqrt(2)) and (0, -asqrt(2))

F = (0, asqrt(2)) F_prime = (0, -asqrt(2))

Distances PF and PF'

PF = sqrt((P[0] - F[0])**2 + (P[1] - F[1])**2) PF_prime = sqrt((P[0] - F_prime[0])**2 + (P[1] - F_prime[1])**2)

Product of distances

product = PF * PF_prime

Set the product equal to 8 and solve for a

equation = Eq(product, 8) solution = solve(equation, a) solution[0].simplify() [/PYTHON]

[THOUGHT] The Python output indicates an error, but the problem can be solved analytically. The hyperbola is (\dfrac{y^2}{a^2} - \dfrac{x^2}{a^2} = 1), and the tangent at ((a, \sqrt{2}a)) has slope (1). The tangent line is (y = x + \sqrt{2}a), so (P = (0, \sqrt{2}a)). The foci are at ((0, \pm a\sqrt{2})). The distances are (\overline{PF} = 2a) and (\overline{PF'} = 2a), so their product is (4a^2 = 8). Thus, (a^2 = 2), and (p + q = 2 + 1 = 3). [/THOUGHT]

\boxed{3}