过抛物线y2=2px的焦点的直线x-my+m=0与抛物线交于A,B两点,且△OAB的面积为2√2,则m^6+m^4=?
2个回答
由题意,可知该抛物线的焦点为(p/2,0),它过直线,代入直线方程,可知：
p/2+m=0--->m=-p/2
∴直线方程变为：y=-2x/p+1
A,B两点是直线与抛物线的交点,∴它们的坐标都满足这两个方程.
∴ (-2x/p+1)^2=2px
---->4x^2/p^2-4x/p+1=2px
---->4x^2/p^2-(4/p+2p)x+1=0
∴△=(4/p+2p)^2-16/p^2=4p^2+16>0
∴方程的解x1=(4/p+2p-√(4p^2+16))/(8/p^2),
x2=(4/p+2p+√(4p^2+16))/(8/p^2);
代入直线方程,可知：y1=1-(4/p+2p-√(4p^2+16))/(4/p),
y2=1-(4/p+2p+√(4p^2+16))/(4/p),
△OAB的面积可分为△OAP与△OBP的面积之和,而△OAP与△OBP若以OP为公共底,则其高即为A,B两点的y轴坐标的绝对值,由图像可知,A,B两点必在x轴两侧,所以它们的y轴坐标必有一个为正,一个为负,比较y1,y2可知y20
∴△OAP与△OBP的面积之和为：
S=1/2 * p/2 *(y1-y2)=p/4 * 2√(4p^2+16)/(4/p)
=p^2/8 *√(4p^2+16)
=2√2
---->p^2√(p^2+4)=8√2
---->p^2=4
∵m=-p/2---->m^2=p^2/4=1
∴m^6+m^4=1^3+1^2=1+1=2
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