Question

已知过抛物线y^2=2P×(P>0)的焦点F的直线与抛物线相交于M、N两点,自M、N向准线l作垂线,垂足分别为Ml、N1

Answer 1

不影响结果,不妨设M在x轴上方,N在x轴下方

设M(m²/(2p),m),N(n²/(2p),n)

AB的方程:(y - n)/(m - n) = [x - n²/(2p)]/[m²/(2p) - n²/(2p)]

(m+n)y - mn = 2px

AB过F(p/2,0):p² = -mn (i)

准线x = -p/2

M₁(-p/2,m),N₁(-p/2,n)

S₁ = (1/2)M₁M*M的纵坐标 = (1/2)m[m²/(2p) + p/2]

S₃ = (1/2)N₁N*|N的纵坐标| = (1/2)(-n)[n²/(2p) + p/2]

S₂ = (1/2)N₁M₁*F和准线的距离 = (1/2)(m- n)(p/2 + p/2) = p(m - n)/2

S₂² = p²(m - n)²/4 (ii)

4S₁S₃ = 4(1/4)m[m²/(2p) + p/2](-n)[n²/(2p) + p/2]

= -mn[m²/(2p) + p/2][n²/(2p) + p/2]

= p²[m²/(2p) + p/2][n²/(2p) + p/2]

= (1/4)(m² + p²)(n² + p²)

= (1/4)[m²n² + (m² + n²)p² + p⁴]

= (1/4)[(-p²)² + (m² + n²)p² + p⁴]

= (p²/4)(2p² + m² + n²)

= (p²/4)(-2mn + m² + n²)

= p²(m - n)²/4

S₂² = 4S₁S₃