Answer:
Option A,C,D
Explanation:
Here, x+iy=1a+ibt×a−ibta−ibt
∴ x+iy=a−ibta2+b2t2
Let a≠ 0,b≠ 0
∴ x=aa2+b2t2 and y=−bta2+b2t2
⇒ yx=−bta⇒t=aybx
On putting x=aa2+b2t2, we get
x(a2+b2.a2y2b2x2)=a
⇒a2(x2+y2)=ax
or x2+y2−xa=0 .........(i)
or (x−12a)2+y2=14a2
∴ Option (a) is correct.
For a≠ 0, b=0
x+iy=1a⇒x=1a,y=0
⇒ z lies on X-axis
∴ Option (c) is correct.
For a=0, b≠ 0,
x+iy=1ibt
⇒ x=0, y=−1bt
⇒ z lies on Y-axis
∴ Option (d) is correct