gaobaopeng
春芽
共回答了14个问题采纳率:92.9% 举报
C1:x^2=2py(p>0),焦点坐标是(0,p/2);准线的方程是y= -p/2;
焦点在C2:y=x^2/2+1上,则p/2=1,p=2,则C1:x^2=4y,准线的方程是y= -1;
P(p,p^2/4),M(m,m^2/2+1),N(n,n^2/2+1),mn;
C2在点M处的切线方程:y-(m^2/2+1)=m(x-m)
C2在点N处的切线方程:y-(n^2/2+1)=n(x-n)
p^2/4-(m^2/2+1)=m(p-m)=mp-m^2,p^2/4=mp-m^2/2+1,(p-2m)^2=4+2m^2
p^2/4-(n^2/2+1)=n(p-n)=np-n^2,p^2/4=np-n^2/2+1,(p-2n)^2=4+2n^2
p=(m+n)/2
m^2-6mn+n^2=16,
MN:(x-n)/(m-n)=(y-n^2/2-1)/(m^2/2-n^2/2),(x-n)(m+n)=2y-n^2-2,
(m+n)x-2y+2-mn=0
设直线 L 的方程为Ax+By+C=0,点 P 的坐标为(Xo,Yo),则点 P 到直线 L 的距离为:
d=│AXo+BYo+C│ / √(A²+B²).
d=|(m+n)p-p^2/2+2-mn|/√[(m+n)^2+4]
=|3(m+n)^2/8+2-mn|/√[(m+n)^2+4]
=|3n^2-2mn+3m^2+16|/[8√(m^2+n^2+2mn+4)],{m^2-6mn+n^2=16,m^2+n^2=16+6mn}
=|64+16mn|/[8√(20+8mn)]
=|4+mn|/√(5+2mn)=|(5+2mn)/2+3/2|/√(5+2mn)
=(1/2)√(5+2mn)+(3/2)/√(5+2mn)
>=2*√[(1/2)√(5+2mn)*(3/2)/√(5+2mn)]
=√3
1年前
8