2p=4
p=2
F(1,0)
MF=3
设直线方程ky=x-4 x=ky+4
带入抛物线方程
y²=4(ky+4)
y²-4ky-16=0
y1+y2=4k
y1y2=-16
SΔAFB=1/2*3*|y1-y2|
S²=9/4(y1-y2)²=9/4[(y1+y2)²-4y1y2]=9/4(16K²+64)≥9/4*64=144
也就是k=0的时候
Smin=12 此时y1=4 y2=-4 x1=x2=4
算错了好几次,真对不起
过定点M(4,0)作直线l,交抛物线y2=4x于A,B两点,F是抛物线的焦点,求三角形AFB面积的最小值
过定点M(4,0)作直线l,交抛物线y2=4x于A,B两点,F是抛物线的焦点,求三角形AFB面积的最小值
数学人气:431 ℃时间:2019-10-19 21:18:15
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