QJEEM20Ma9dd3: JEE Main 2020 Mathematics - Class 11 JEEM20Ma9dd3 Solution: Let , z = x + iy and k > 0. If the curve represented by Re(u) + …

Given, z = x + iy

and $u = {{2z + i} \over {z - ki}}$

$= {{2(x + iy) + i} \over {(x + iy) - ki}}$

$= {{2x + i(2y + 1)} \over {x + i(y - k)}} \times {{x - i(y - k)} \over {x - i(y - k)}}$

$= {{2{x^2} + (2y + 1)(y - k) + i(2xy + x - 2xy + 2kx)} \over {{x^2} + {{(y - k)}^2}}}$

Given,

${\mathop{\rm Re}\nolimits} (u) + {\mathop{\rm Im}\nolimits} (u) = 1$

$\Rightarrow {{2{x^2} + (2y + 1)(y - k)} \over {{x^2} + {{(y - k)}^2}}} + {{x + 2kx} \over {{x^2} + {{(y - k)}^2}}} = 1$

$\Rightarrow 2{x^2} + (2y + 1)(y - k) + x + 2kx = {x^2} + {(y - k)^2}$

This curve intersect the y-axis at point P and Q, so at point P and Q x = 0

Putting x = 0 at the above equation,

$\therefore$ $(2y + 1)(y - k) = {(y - k)^2}$

$\Rightarrow 2{y^2} + y - 2yk - k = {y^2} + {k^2} - 2ky$

$\Rightarrow {y^2} + y - (k + {k^2}) = 0$

Let roots of this quadratic equation y₁ and y₂

$\therefore$ Point P (0, y₁) and Q (0, y₂)

and ${y_{_1}} + {y_2} = 1$ , ${y_1}{y_2} = - k - {k^2}$

$\therefore$ ${({y_1} - {y_2})^2} = {({y_1} + {y_2})^2} - 4{y_1}{y_2}$

$= 1 + 4k + 4{k^2}$

$\Rightarrow |{y_1} - {y_2}| = \sqrt {1 + 4k + 4{k^2}}$

Given, PQ = 5

$\Rightarrow |{y_1} - {y_2}| = 5$

$\Rightarrow \sqrt {1 + 4k + 4{k^2}} = 5$

$\Rightarrow {k^2} + k - 6 = 0$

$\Rightarrow k = - 3,\,2$

So, k = 2 (Given k > 0)