Question

已知过抛物线y2=2px(p>0)的焦点的弦长为5/2p,求弦所在直线的方程 谢谢

Answer 1

焦点F(p/2, 0)
设弦所在直线斜率为k, 方程为y = k(x - p/2), x = y/k + p/2
y² = 2p(y/k + p/2) = 2py/k + p²
ky² -2py - kp² = 0
y = [p ± p√(k² + 1)]/k
x = [p ± p√(k² + 1)]/k² + p/2
弦长² = 25p²/4 = (x1 - x2)² + (y1 - y2)² = 4p²(k² + 1)²/k⁴
k = ±√6/3
直线的方程y = (±√6/3)(x - p/2)

追问

为什么y＝k﹙x－p／2﹚

追答

标准形式为y2=2px时，焦点为(p/2, 0), 准线为x = -p/2