初三数学抛物线

Question

如题所述

Answer 1

(1)
y = x/2 + 2 = x²/4, x² - 2x - 8 = (x + 2)(x - 4)＝0
x = -2, A(-2, 1)
x = 4, B(4, 4)

(2)
令B(t, t²/4), BD的斜率为m, 则BD的方程为y - t²/4 = m(x - t)
与抛物线联立: x²/4 - t²/4 = m(x - t)
x² - 4mx + t(4m - t) = 0
Δ = (-4m)² - 4t(4m - t) = 4(2m - t) = 0, m = t/2
BD: y - t²/4 = (t/2)(x - t)
令x = 0, y = -t²/4, D(0, -t²/4)
C(0, b), E(0, 2b)
AB的方程: (y - b)/(t²/4 - 0) = (x - 0)/(t - 0)
y = (t/4)x + b
与y = kx + b比较, k = t/4, t = 4k
B(4k, 4k²), D(0, -4k²)
BC² = (4k - 0)² + (4k² - b)² = CD² = (-4k² - b)²
k²(b - 1) = 0
b = 1
E(0, 2), y = x²/4 = 2, x = -2√2
F(-2√2, 0), EF = 2√2

(3)
按(2)中的办法, 得BD: y - 4 = 2(x - 4), y = 2x - 4
P(p, 2p - 4), p < 4
N(p, p²/4), Q(p, p/2 +2)
AN的斜率为u = (1 - p²/4)/(-2 - p) = (p - 2)/4
PM的斜率也是u, 其方程为: y - (2p - 4) = [(p - 2)/4](x - p)
与AB联立，得M(p-6, (p - 2)/2)
MQ² = (p - 6 - p)² + [(p - 2)/2 - p/2 - 2]² = 45
MQ = 3√5