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Q.
Let $u =\frac{2 z + i }{ z - ki }, z = x + iy$ and $k >0 .$ If the curve
represented by $Re( u )+Im( u )=1$ intersects the y-axis at the points $P$ and $Q$ where $P Q=5,$ then
the value of $k$ is :
JEE MainJEE Main 2020Complex Numbers and Quadratic Equations
Solution:
$u =\frac{2 z + i }{ z - ki }$
$=\frac{2 x ^{2}+(2 y +1)( y - k )}{ x ^{2}+( y - k )^{2}}+ i \frac{( x (2 y +1)-2 x ( y - k ))}{ x ^{2}+( y - k )^{2}}$
Since $Re( u )+Im( u )=1$
$\Rightarrow 2 x ^{2}+(2 y +1)( y - k )+ x (2 y +1)-2 x ( y - k )$
$=x^{2}+( y - k )^{2}$
$\because PQ = 5$
$\Rightarrow \mid y _{1}- y _{2} | =5$
$ \Rightarrow k ^{2}+ k -6=0$
$\Rightarrow k =-3,2$
$So , k =2( k >0)$
