设抛物线C的顶点在原点,焦点F在x轴上,
1个回答
抛物线C上横坐标为3的点到C的准线的距离等于4.
即有x+p/2=4,3+p/2=4,p=2
即抛物线方程是y^2=4x.
F坐标是（1,0）,则设AB方程是x=my+1.
代入得到:y^2=4my+4,y^2-4my-4=0
y1+y2=4m,y1y2=-4
x1x2=y1^2y2^2/16=1
x1+x2=m(y1+y2)+2=4m^2+2
NA^2*NB^2=[(3-x1)^2+y1^2][(3-x2)^2+y2^2]
=(9-6x1+x1^2+4x1)(9-6x2+x2^2+4x2)
=(9-2x1+x1^2)(9-2x2+x2^2)
=81-18x2+9x2^2-18x1+4x1x2-2x1x2^2+9x1^2-2x1^2x2+x1^2x2^2
=81-18(x1+x2)+9(x1^2+x2^2)+4*1-2x1x2(x1+x2)+1
=86-18(4m^2+2)+9[(4m^2+2)^2-2]
=86-72m^2-36+9(16m^4+16m^2+4)-18
=144m^4+72m^2+68
=144(m^2+1/4)^2+68-9
故当m^2=0时有最小值是：68
即NA*NB的最小值是：根号68=2根号17
相关问题
