Question

If the equation $|z-a|+|z-b|=3$ represents an ellipse, and $a,b\in C,$ where a is fixed, then find the locus of b.

• A. $|b-a|>3$
• B. $|b-a|<3$
• C. $|b-2a|>3$
• D. $|b-2a|<3$

Answer 1

Answer: (B)

$|b-a|<3$

Let $z=x+iy$
$|z-a|+|z-b|=3$
$\sqrt{(x-a{)}^2+y^2}+\sqrt{(x-b{)}^2+y^2}=3$
$\sqrt{(x-a{)}^2+y^2}=3-\sqrt{(x-b{)}^2+y^2}$
$(x-a{)}^2+y^2=9+(x-b{)}^2+y^2-6\sqrt{(x-b{)}^2+y^2}$
$6\sqrt{(x-b{)}^2+y^2}=9+(x-b{)}^2-(x-a{)}^2$
$6\sqrt{(x-b{)}^2+y^2}=9+(a-b)(2x-(a+b\left)\right)$
$6\sqrt{(x-b{)}^2+y^2}=(9+a^2-b^2)+2(a-b)x$
$36y^2+36(x-b{)}^2=4(a-b{)}^2{x}^2+18(a-b)(9+a^2-b^2)x+\lambda$
Therefore coefficient of $x^2$ will be
$36-4(a-b{)}^2$.
Now for the above equation to represent an ellipse
$36-4(a-b)6^2>0$
$4(a-b{)}^2<36$
$(a-b{)}^2<9$
$|a-b|<3$