Question

高三数学第二问

Answer 1

c1:x^2=2py,c2:x^2+y^2=8=y^2+2py,(y+p)^2=8+p^2,y=+-√(8+p^2)-p
y>=0,y=√(8+p^2)-p,x=+-√[√(8+p^2)-p]
c1,c2都关于y轴对称，所以y^2=8-(4/2)^2=4,y=2,所以(2+p)^2=8+p^2,p=1
c1:y=x^2/2,A(xa,xa^2/2),B(xb,xb^2/2)
c1:y'=x,
AP:y=xa(x-xa)+xa^2/2=xax+xa^2/2,
BP:y=xb(x-xb)+xb^2/2=xbx+xb^2/2;
P:x=(xa^2/2-xb^2/2)/(xb-xa)=-(xa+xb)/2,y=xax+xa^2/2=-xaxb/2
OP:x^2+y^2=(xa+xb)^2/4+xa^2xb^2/4<8,xa^2+xb^2+2xaxb+xa^2xb^2<32
32>xa^2xb^2+2xaxb+xa^2+xb^2>=xa^2xb^2+4xaxb
(xaxb+2)^2<36,-8<xaxb<4
AB:y=(yb-ya)(x-xa)/(xb-xa)+ya=(1/2)(xb^2-xa^2)(x-xa)/(xb-xa)+xa^2/2
=(xb+xa)(x-xa)/2+xa^2/2=(xa+xb)x/2-xaxb/2
(yd-yc)/(xd-xc)=(xa+xb)/2
c2:x^2+[(xa+xb)x/2-xaxb/2]^2=8
x^2[1+(xa+xb)^2/4]-x(xa+xb)xaxb/2+xa^2xb^2/4-8=0
(xd-xc)^2={[(xa+xb)xaxb/2]^2-4[1+(xa+xb)^2/4][xa^2xb^2/4-8]}/[1+(xa+xb)^2/4]^2
|CD|^2=(yd-yc)^2+(xd-xc)^2=[1+(xa+xb)^2/4](xd-xc)^2
={[(xa+xb)xaxb/2]^2-4[1+(xa+xb)^2/4][xa^2xb^2/4-8]}/[1+(xa+xb)^2/4]
={[(xa+xb)xaxb/2]^2/[1+(xa+xb)^2/4]-[xa^2xb^2-32]}
|AB|^2=(yb-ya)^2+(xb-xa)^2=(xb^2-xa^2)^2/4+(xb-xa)^2
=(xb-xa)^2[(xa+xb)^2/4+1]
(|AB||CD|)^2=(xb-xa)^2[(xa+xb)xaxb/2]^2-[1+(xa+xb)^2/4][xa^2xb^2-32]
=(xb-xa)^2(xa+xb)^2xa^2xb^2/4-xa^2xb^2+32+8(xa+xb)^2-(xa+xb)^2xa^2xb^2/4
=[(xb^2-xa^2)^2-(xa+xb)^2]xa^2xb^2/4-xa^2xb^2+32+8xa^2+8xb^2+16xaxb
先吃饭了