Problem Analysis #geometry_26

Recorded Python call:

import sympy as sp

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

# Hyperbola: x^2/a^2 - y^2/a^2 = -1  (vertical hyperbola, A^2 = B^2 = a^2)
# Foci on y-axis: c^2 = a^2 + a^2 = 2a^2 -> c = sqrt2 * a
c = sp.sqrt(2)*a
F = (0, c); Fp = (0, -c)

# Point (a, sqrt2 a) on hyperbola: check
on_curve = sp.simplify(a**2/a**2 - (sp.sqrt(2)*a)**2/a**2 + 1)
print("on curve check (=0):", on_curve)

# Tangent at (x1,y1): x*x1/a^2 - y*y1/a^2 = -1  ->  x - sqrt2 y = -a
x1, y1 = a, sp.sqrt(2)*a
# Intersection with y-axis (x=0): -y1*y/a^2 = -1 -> y = a^2/y1 = a/sqrt2
yP = sp.simplify(a**2/y1)
print("P_y =", yP)

# Distances along y-axis
PF  = sp.simplify(c - yP)
PFp = sp.simplify(yP + c)
print("PF =", PF, "  PF' =", PFp)

product = sp.simplify(PF*PFp)
print("PF*PF' =", product)

sol = sp.solve(sp.Eq(product, 8), a)
print("a =", sol, "-> rationalized:", [sp.simplify(s) for s in sol])

풀이 요약

• 쌍곡선 $\frac{x^2}{a^2}-\frac{y^2}{a^2}=-1$은 y축을 실축으로 하며, $c^2=a^2+a^2=2a^2$이므로 초점은 $F(0,\sqrt2 a)$, $F'(0,-\sqrt2 a)$.
• 점 $(a,\sqrt2 a)$에서의 접선: $\frac{x\cdot a}{a^2}-\frac{y\cdot\sqrt2 a}{a^2}=-1 \Rightarrow x-\sqrt2 y=-a$. y축과의 교점은 $P\big(0,\frac{a}{\sqrt2}\big)=\big(0,\frac{\sqrt2}{2}a\big)$.
• y축 위의 거리들: $\overline{PF}=\sqrt2 a-\frac{\sqrt2}{2}a=\frac{\sqrt2}{2}a$, $\overline{PF'}=\frac{\sqrt2}{2}a+\sqrt2 a=\frac{3\sqrt2}{2}a$.

Python 검증 (sympy): 접선 조건, 접선 방정식, 거리식을 기호로 계산한 결과 $\overline{PF}\cdot\overline{PF'}=\frac{3a^2}{2}$, $\frac{3a^2}{2}=8 \Rightarrow a=\frac{4\sqrt3}{3}$ (정확 해) — 이는 선택지 ②에 해당합니다.

$\boxed{2}$