QJEEAdv16Ma4c23: JEE Advanced 2016 Mathematics - Class 11 JEEAdv16Ma4c23 Solution: Let . Suppose , where . Ifz = x + iy and z S, then (x, y) lies on

Here, $x + iy = {1 \over {a + ibt}} \times {{a - ibt} \over {a - ibt}}$

$\therefore$ $x + iy = {{a - ibt} \over {{a^2} + {b^2}{t^2}}}$

Let a $\ne$ 0, b $\ne$ 0

$\therefore$ $x = {a \over {{a^2} + {b^2}{t^2}}}$ and $y = {{ - bt} \over {{a^2} + {b^2}{t^2}}}$

$\Rightarrow {y \over x} = {{ - bt} \over a} \Rightarrow t = {{ay} \over {bx}}$

On putting $x = {a \over {{a^2} + {b^2}{t^2}}}$ , we get

$x\left( {{a^2} + {b^2}\,.\,{{{a^2}{y^2}} \over {{b^2}{x^2}}}} \right) = a$

$\Rightarrow {a^2}({x^2} + {y^2}) = ax$

or, ${x^2} + {y^2} - {x \over a} = 0$ ...... (i)

or, ${\left( {x - {1 \over {2a}}} \right)^2} + {y^2} = {1 \over {4{a^2}}}$

$\therefore$ Option (a) is correct.

For a $\ne$ 0 and b = 0,

$x + iy = {1 \over a} \Rightarrow x = {1 \over a},\,y = 0$

$\Rightarrow$ z lies on X-axis.

$\therefore$ Option (c) is correct.

For a = 0 and b $\ne$ 0,

$x + iy = {1 \over {ibt}}$

$\Rightarrow x = 0,\,y = - {1 \over {bt}}$

$\Rightarrow$ z lies on Y-axis.

$\therefore$ Option (d) is correct.