Question

The equation $|z+i|-|z-i|=k$ represents a hyperbola, if

• A. $k\in (-2,2)$
• B. $k\in (-2,0)$
• C. $k\in \{0,2\}$
• D. $k\in \{-2,2\}$

Answer 1

The correct option is A $k\in (-2,2)$

Let $z=x+iy$
Therefore
$|z-i|+|z+i|=k$

$\sqrt{x^2+(y-1{)}^2}+\sqrt{x^2+(y+1{)}^2}=k$
$\sqrt{x^2+(y-1{)}^2}=k-\sqrt{x^2+(y+1{)}^2}$
$x^2+(y-1{)}^2=k^2+x^2+(y+1{)}^2-2k\sqrt{x^2+(y+1{)}^2}$
$2k\sqrt{x^2+(y+1{)}^2}=k^2+(y+1{)}^2-(y-1{)}^2$
$2k\sqrt{x^2+(y+1{)}^2}=k^2+4y$
$4k^2(x^2+(y+1{)}^2)=k^4+16y^2+8k^{2}y$
$4k^2{x}^2+4k^2{y}^2+4k^{2}y+4k^2=k^4+16y^2+8k^{2}y$
$4k^2{x}^2+y^2(4k^2-16)-4k^{2}y+4k^2-k^4=0$
Therefore it represents a hyperbola if
$4k^2-16<0$
Or
$4k^2<16$
$k^2<4$
$|k|<2$
$-2<k<2$