Question

1.已知z1,z2∈C,且5z1^2+2z1^2=z1z2,z1+z2=1,则2z1-z2=

Answer 1

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