Problem Analysis #45

[PYTHON] from sympy import symbols, sqrt, solve, Abs, simplify

def solve_hyperbola(): a = symbols('a', positive=True)

# Hyperbola: x^2/a^2 - y^2/a^2 = -1 => y^2/a^2 - x^2/a^2 = 1
# This is a vertical hyperbola.
# c^2 = a^2 + a^2 = 2a^2 => c = a*sqrt(2)
c = a * sqrt(2)

F = (0, c)
F_prime = (0, -c)

# Point of tangency T = (a, sqrt(2)*a)
x0 = a
y0 = sqrt(2) * a

# Tangent equation for y^2/a^2 - x^2/a^2 = 1 at (x0, y0) is y*y0/a^2 - x*x0/a^2 = 1
# y * (sqrt(2)a) / a^2 - x * a / a^2 = 1
# sqrt(2)*y/a - x/a = 1
# sqrt(2)*y - x = a

# Find P (intersection with y-axis, so x=0)
# sqrt(2)*y_P - 0 = a => y_P = a/sqrt(2)
y_P = a / sqrt(2)
P = (0, y_P)

# Calculate distances PF and PF'
# Since P, F, F' are on y-axis, distance is absolute difference of y-coordinates
dist_PF = Abs(y_P - c)
dist_PF_prime = Abs(y_P - (-c))

# Condition: PF * PF' = 8
equation = dist_PF * dist_PF_prime - 8

# Solve for a
solutions = solve(equation, a)
print(f"Solutions for a: {solutions}")

# We need a^2
vals = []
for sol in solutions:
    val_sq = sol**2
    print(f"a={sol}, a^2={val_sq}")
    vals.append(val_sq)
    
return vals

results = solve_hyperbola() print(results) [/PYTHON]

[THOUGHT] The problem asks for $p+q$ where $a^2 = \frac{p}{q}$. From the Python code, we found that the condition $\overline{PF} \times \overline{PF'} = 8$ leads to the equation $3a^2/2 = 8$. Solving for $a^2$, we get $a^2 = \frac{16}{3}$. Thus, $p = 16$ and $q = 3$. These are coprime integers since $\text{gcd}(16, 3) = 1$. We need to compute $p+q = 16 + 3 = 19$. [/THOUGHT]

\boxed{19}