AP EAMCET201920 Apr 2019Morning ShiftMathematicsComplex NumberActual
If (z=x+i y, x, y R,(x, y) (0,-4) ) and Arg ( ( 2 z-3 z+4 i )= 4 ), then the locus of (z ) is
Options
- A(2 x^2+2 y^2+5 x+5 y-12=0 )
- B(2 x^2-3 x y+y^2+5 x+y-12=0 )
- C(2 x^2+3 x y+y^2+5 x+y+12=0 )
- D(2 x^2+2 y^2-11 x+7 y-12=0 )
Correct answer
A. (2 x^2+2 y^2+5 x+5 y-12=0 )
Step-by-step solution
( aligned & For z=x+i y, x, y R,(x, y) (0,-4) & 2 z-3 z+4 i = (2 x-3)+2 i y x+i(y+4) x-i(y+4) x-i(y+4) & = (2 x^2-3 x+2 y^2+8 y )+i(2 x y-2 x y+3 y-8 x+12) x^2+(y+4)^2 & = (2 x^2-3 x+2 y^2+8 y )+i(12+3 y-8 x) x^2+(y+4)^2 & So, ( 2 z-3 z+4 i )= ⁻¹ ( 12+3 y-8 x 2 x^2-3 x+2 y^2+8 y )= 4 (given) & 12+3 y-8 x 2 x^2-3 x+2 y^2+8 y =1 & 2 x^2+2 y^2+5 x+5 y-12=0 aligned ) Hence, option (1) is correct.