直线l与抛物线y∧2=2px交于a(x1,y1),b(x2,y2)两点,若y1y2=-p∧2,求证:直线l过抛物线的焦点f

问题描述:

直线l与抛物线y∧2=2px交于a(x1,y1),b(x2,y2)两点,若y1y2=-p∧2,求证:直线l过抛物线的焦点f

设直线l的方程为y = kx + b,x = (y - b)/k
代入y² = 2px:
y² = 2p(y-b)/k
ky² - 2py + 2pb = 0
y1*y2 = 2pb/k = -p²
b = -pk/2
l的方程为y = kx -pk/2 = k(x - p/2)
显然,l过点(p/2,0),此外为该抛物线的焦点.