斜率为1的直线经过抛物线y^2=4x的焦点,且与抛物线相交于A,B两点,则绝对值AB等于____
1个回答
展开全部
y^2=4x = 2*2x = 2px
p = 2
焦点F(p/2, 0), 即(1, 0)
设直线方程为 y = x + b
过F(1, 0), 0 = 1 + b, b = -1
y = x -1
(x-1)^2 = 4x
x^2 -6x + 1 = 0
x1 = 3 - 2√2, x2 = 3+2√2
y1 = 2 - 2√2, y2 = 2+2√2
|AB| = √[(x2 -x1)^2 + (y2 - y1)^2] = √[(4√2)^2 + (4√2)^2] = 8
p = 2
焦点F(p/2, 0), 即(1, 0)
设直线方程为 y = x + b
过F(1, 0), 0 = 1 + b, b = -1
y = x -1
(x-1)^2 = 4x
x^2 -6x + 1 = 0
x1 = 3 - 2√2, x2 = 3+2√2
y1 = 2 - 2√2, y2 = 2+2√2
|AB| = √[(x2 -x1)^2 + (y2 - y1)^2] = √[(4√2)^2 + (4√2)^2] = 8
来自:求助得到的回答
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询