迷信中药
幼苗
共回答了22个问题采纳率:81.8% 举报
1.5z1^2+2z1^2=z1z2,应该有一个是z2^2
2.设P(x1,y1) x1^2-y1^2=k y1^2=x1^2-k
a^2=b^2=k
c^2=2k c=√(2k) F1(-√(2k),0) F2(√(2k),0)
|PF1|=√[(x1+√(2k)^2)+y1^2]=√[x1^2+2x1√(2k)+2k+y1^2]=√[2x1^2+k+2x1√(2k)]
|PF2|=√[(x1-√(2k)^2)+y1^2]=√[x1^2-2x1√(2k)+2k+y1^2]=√[2x1^2+k-2x1√(2k)]
|PF1|*|PF2|=√[(2x1^2+k+2x1√(2k))(2x1^2+k-2x1√(2k))]
=√[(2x1^2+k)^2-8x1^2k]
=√[4x1^4-4x1^2k+k^2]
=2x1^2-k
=x1^2+y1^2
1年前
追问
6
举报
迷信中药
z1+z2=1 z2=1-z1 5z1^2+2(1-z1)^2-z1(1-z2)=0 8z1^2-5z1+2=0 z1=(5±√-39)/16 z1=(5+√39i )/16 2z1-z2=3z1-1=(-1+3√39i)/16 z1=(5-√39i )/16 2z1-z2=3z1-1=(-1-3√39i)/16