Sorry. I misread it.
\hspace{-16}$If $\mathbf{a,b\in \mathbb{R}}$ and $\mathbf{a\neq b}$. Then Locus of all Complex no. $\mathbf{z}$ which\\\\ Satisfy the equation. $\mathbf{\mid z-a \mid^2-\mid z-b \mid^2=1}$
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3 Answers
Aditya Bhutra
·2012-03-23 07:10:56
putting in z= x+iy and simplifying we get
x= 1+b2-a22(b-a)
thus locus should be Re(z) = 1+b2-a22(b-a)