高三数学圆锥曲线题，请高手解答，多谢！

Question

如题所述

Answer 1

抛物线x^2=4y的焦点坐标是F(0,1),即有椭圆的c=1.又有A(0,2),即有a=2, b^2=a^2-c^2=4-1=3
故椭圆E方程是y^2/4+x^2/3=1.
设过F(0,1)的直线方程是y=kx+1.代入到抛物线中有x^2=4(kx+1)
即有x^2-4kx-4=0
设C坐标是(x1,y1),D(x2,y2)
y=x^2/4, y'=x/2,则有L1的斜率k1=x1/2, L2的斜率k2=x2/2
故有k1*k2=x1x2/4=-4/4=-1
故有L1和L2垂直.
y=kx+1代入到椭圆中有(kx+1)^2/4+x^2/3=1,即有4x^2+3(k^2x^2+2kx+1)=12
(4+3k^2)x^2+6kx-9=0
设M坐标是(x3,y3),N(x4,y4),则有x3+x4=-6k/(4+3k^2),x3x4=-9/(4+3k^2)
又有S(AMN)=1/2AF*|X3-X4|=1/2*1*|x3-x4|=1/2根号[(x3+x4)^2-4x3x4]=1/2根号[36k^2/(4+3k^2)+36/(4+3k^2)]=1/2根号(36k^2+144+108k^2)/(4+3k^2)^2=1/2*12/(4+3k^2)*根号(k^2+1)=6根号(k^2+1)/(3k^2+4)
设t=根号(k^2+1)>=1,则有S=6t/(3t^2+1)=6/(3t+1/t)<=6/ (3+1)=3/2
即当根号(k^2+1)=1,即k=0时取得最大值是3/2.追问

老师还有一个