Let $a, b \in \mathbb{R}$ and $a^{2} + b^{2} \neq 0$. Suppose $S = \left\{ z \in \mathbb{C} : Z = \frac{1}{a+ibt}, \in \mathbb{R}^{+}, t \neq 0 \right\}$, where $i = \sqrt{-1}$. If $z = x + iy$ and $z \in S$, then $(x, y)$ lies on

Let a, b ∈ R and a² + b² ≠ 0. Suppose S = { z ∈ C : Z = 1/(a+ibt), ∈ R⁺, t ≠ 0 }, where i = √-1. If z = x + iy and z ∈ S, then (x, y) lies on $\text{Given: } a, b \in \mathbb{R}, \quad a^{2} + b^{2} \neq 0$ $S = \left\{ z \in \mathbb{C} : z = \dfrac{1}{a+ibt} \in \mathbb{R}^{+},\ t \neq 0 \right\}$ $i = \sqrt{-1}, \quad z = x + iy$ $\text{To find: the locus of } (x, y), \text{ where } z = x + iy \in S$ $\text{Choose the correct curve from the four options}$ $1.\ \text{rationalise } \dfrac{1}{a+ibt} \text{ into } x + iy$ $2.\ \text{separate } x \text{ and } y, \text{ then eliminate } t$ $3.\ \text{check special cases } b = 0 \text{ and } a = 0$ $z = \dfrac{1}{a+ibt} = \dfrac{a-ibt}{(a+ibt)(a-ibt)}$ $z = \dfrac{a - ibt}{a^{2} + b^{2}t^{2}}$ $x = \dfrac{a}{a^{2}+b^{2}t^{2}}$ $y = \dfrac{-bt}{a^{2}+b^{2}t^{2}}$ $x^{2} + y^{2} = \dfrac{a^{2} + b^{2}t^{2}}{(a^{2}+b^{2}t^{2})^{2}} = \dfrac{1}{a^{2}+b^{2}t^{2}}$ $x = a \cdot \dfrac{1}{a^{2}+b^{2}t^{2}} = a\,(x^{2}+y^{2})$ $a\,(x^{2}+y^{2}) = x \ \Rightarrow \ x^{2} + y^{2} = \dfrac{x}{a} \quad (a \neq 0)$ $\left(x - \dfrac{1}{2a}\right)^{2} + y^{2} = \left(\dfrac{1}{2a}\right)^{2}$ $\text{centre } \left(\dfrac{1}{2a}, 0\right), \quad \text{radius } \left|\dfrac{1}{2a}\right|$ $a > 0,\ b \neq 0: \text{ radius } \dfrac{1}{2a}, \ \text{centre } \left(\dfrac{1}{2a}, 0\right) \Rightarrow (\text{1})$ $a < 0: \text{ centre } \left(\dfrac{1}{2a}, 0\right) \neq \left(-\dfrac{1}{2a}, 0\right) \Rightarrow (\text{2}) \text{ is wrong}$ $b = 0,\ a \neq 0: z = \dfrac{1}{a} \in \mathbb{R} \Rightarrow y = 0 \Rightarrow (\text{3})$ $a = 0,\ b \neq 0: z = \dfrac{1}{ibt} = -\dfrac{i}{bt} \Rightarrow x = 0 \Rightarrow (\text{4})$ $\text{Correct options: } (\text{1}),\ (\text{3}),\ (\text{4})$