Question

已知椭圆c1的离心率为根号3/2，抛物线c2：x^2=4y的焦点在椭圆的顶点上（1）过A(0,1)

的直线l与抛物线c2交于e，f两点，又过e，f分别作抛物线c2的切线并交于点M，求证：以ef为直径的园过点m

Answer 1

(1) e² = c²/a² = (a² - c²)/a² = (4 - b²)/4 = 3/4, b = 1
C2为开口向上的抛物线，焦点只能是椭圆的上顶点(0, 1), p/2 = 1, p = 2
C2: x² = 4y

(2)F(0, 1/2), p/2 = 1/2, p = 1
x² = 2y, y = x²/2
设P(u, u²/2)
y' = x
过点p的切线: y - u²/2 = u(x - u), y = ux - u²/2
代入x²/4 + y²= 1
(4u² + 1)x² - 4u³x + u⁴ - 4 = 0
x₁ + x₂ = 4u³/(4u² + 1)
x₁x₂ = (u⁴ - 4)/(4u² + 1)
y₁,₂ = ux₁,₂ - u²/2
设OA, OB的斜率分别为m, n
mn = -1 = y₁y₂/(x₁x₂)
-x₁x₂ = y₁y₂
-(u⁴ - 4)/(4u² + 1) = (ux₁ - u²/2)(ux₂ - u²/2) = u²x₁x₂ - (u³/2)(x₁ + x₂) + u⁴/4
= u²(u⁴ - 4)/(4u² + 1) - (u³/2)*4u³/(4u² + 1) + u⁴/4
5u⁴ - 16u² - 16 = 0
(5u² + 4)(u² - 4) = 0
u² = 4
u = ±2
P(±2, 2)