Question

If $z= x +iy , x,y \in R , (x,y) \ne (0, -4)$ and $Arg \left( \frac{2z-3}{z+4i}\right) = \frac{\pi}{4}$ , then the locus of $z$ is

• A. $2x^2 + 2y^2 + 5x + 5y - 12 = 0$
• B. $2x^2 - 3xy + y^2 + 5x + 5y - 12 = 0$
• C. $2x^2 + 3xy + y^2 + 5x + 5y - 12 = 0$
• D. $2x^2 + 2y^2 - 11 x + 7y - 12 = 0$

Answer 1

Answer: (A)

$2x^2 + 2y^2 + 5x + 5y - 12 = 0$

Answer 2

For $z=x+i y, x, y \in R,(x, y) \neq(0,-4)$
$\frac{2 z-3}{z+4 i}=\frac{(2 x-3)+2 i y}{x+i(y+4)} \times \frac{x-i(y+4)}{x-i(y+4)}$
$=\frac{\left(2 x^{2}-3 x+2 y^{2}+8 y\right)+i(2 x y-2 x y+3 y-8 x+12}{x^{2}+(y+4)^{2}}$
$=\frac{\left(2 x^{2}-3 x+2 y^{2}+8 y\right)+i(12+3 y-8 x)}{x^{2}+(y+4)^{2}}$
So, $\arg \left(\frac{2 z-3}{z+4 i}\right)=\tan ^{-1}\left(\frac{12+3 y-8 x}{2 x^{2}-3 x+2 y^{2}+8 y}\right)$
$=\frac{\pi}{4}$ (given)
$\Rightarrow \frac{12+3 y-8 x}{2 x^{2}-3 x+2 y^{2}+8 y}=1$
$\Rightarrow 2 x^{2}+2 y^{2}+5 x+5 y-12=0$