Question

If $|z-2-i|=|z||\sin (\frac{\pi }{4}-argz)|$, then locus of $z$ is

• A. A pair of straight line
• B. Circle
• C. Parabola
• D. Ellipse

Answer 1

Answer: (C)

Parabola

$z=x+iy$
$=|z|cos\theta +i|z|sin\theta$
$|z-2-i|$
$=|(x-2)+i(y-1)|$
$=\sqrt{(x-2{)}^2+(y-1{)}^2}$
$=\frac{|z|}{\sqrt{2}}|cos\theta -sin\theta |$
$=\frac{1}{\sqrt{2}}||z|cos\theta -|z|sin\theta |$
$=\frac{1}{\sqrt{2}}|x-y|$
Hence
$(x-2{)}^2+(y-1{)}^2=\frac{(x-y{)}^2}{2}$
$2[x^2-4x+4+y^2-2y+2]=x^2+y^2-2xy$
$x^2+y^2-8x-4y+12=-2xy$
$x^2+y^2+2xy-8x-4y+12=0$
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
$a=1,b=1,c=12,g=-4,f=-2,h=1$
Hence
$abc+2fgh-a{f}^2-b{g}^2-c{h}^2$
$=\left(12\right)+2\left(8\right)-4-16-12$
$=28-32$
$=-4$
$△\ne 0$
$(2h{)}^2-4ab$
$=4-4$
$=0$
Hence it is a parabola.