Question

过点p(0,2)的直线和抛物线y^2=8x交于A,B两点,若线段AB的中点M在直线x=2上,求弦AB的长

Answer 1

y-2=k(x-0)
y=kx+2
代入
k²x²+4kx+4=8x
k²x²+(4k-8)x+4=0
中点在x=2上
即横坐标(x1+x2)/2=2
而x1+x2=-(4k-8)/k²=4
k²+k-2=0
k=-2,k=1

k=-2
则x1+x2=4
x1x2=4/k²=1
(x1-x2)²=(x1+x2)²-4x1x2=12
y=-2x+2
所以(y1-y2)²=(-2x1+2+2x2-2)=4(x1-x2)²=48
所以AB=√[(x1-x2)²-(y1-y2)²]=2√15

k=1
则x1+x2=4
x1x2=4/k²=4
(x1-x2)²=(x1+x2)²-4x1x2=0
y=-2x+2
所以(y1-y2)²=(-2x1+2+2x2-2)=4(x1-x2)²=0
所以AB=√[(x1-x2)²-(y1-y2)²]=0
此时是相切，没有弦

所以AB=2√15

Answer 2

设过P(0,2)的直线是
y-2=kx
代入y^2=8x得
(kx+2)^2=8x
k^2x^2+(4k-8)x+4=0
x1+x2=-(4k-8)/k^2=4
4k^2+4k-8=0
k^2+k-2=0
(k+1)(k-2)=0
k=1,k=-2
当k=1时
x1x2=4/k^2=4/k^2=4,
AB=|x1-x2|*√(1+k^2)
=√(x1-x2)^2*√(1+k^2)
=√[(x1+x2)^2-4x1x2]*√(1+k^2)
=√(16-16)*√2
=0
说明直线与抛物线相切
当k=-2时
x1x2=4/k^2=4/k^2=1,
AB=|x1-x2|*√(1+k^2)
=√(x1-x2)^2*√(1+k^2)
=√[(x1+x2)^2-4x1x2]*√(1+k^2)
=√(16-4)*√5
=2√15
故AB长2√15