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  • Let  and k>0. If the curve represented by <img alt="Re (u)+Im(u)=1" src="http://entrancec

Let u=\frac{2z+i}{z-ki}, z=xiy and k>0. If the curve represented by Re (u)+Im(u)=1 intersects the y-axis at the point P and Q where PQ=5, then the value of k is:
Option: 1 3/2
Option: 2 1/2
Option: 3 4
Option: 4 2

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\\\mathrm{u}=\frac{2 \mathrm{z}+\mathrm{i}}{\mathrm{z}-\mathrm{ki}} \\ =\frac{2 \mathrm{x}^{2}+(2 \mathrm{y}+1)(\mathrm{y}-\mathrm{k})}{\mathrm{x}^{2}+(\mathrm{y}-\mathrm{k})^{2}}+\mathrm{i} \frac{(\mathrm{x}(2 \mathrm{y}+1)-2 \mathrm{x}(\mathrm{y}-\mathrm{k})}{\mathrm{x}^{2}+(\mathrm{y}-\mathrm{k})^{2}}

Since Re(u) + Im(u) = 1

\begin{aligned} \Rightarrow 2 x^{2}+&(2 y+1)(y-k)+x(2 y+1)-2 x(y-k) \\ &=x^{2}+(y-k)^{2} \end{aligned}

Since, the curve intersects the y-axis at the points P and Q

Let P = (0, y1) and Q = (0, y2)

\\\Rightarrow \mathrm{y}^{2}+\mathrm{y}-\mathrm{k}-\mathrm{k}^{2}=0\\y_1+y_2=-1\\y_1y_2=-k-k^2

Given, PQ = 5

\\\Rightarrow\left|\mathrm{y}_{1}-\mathrm{y}_{2}\right|=5 \Rightarrow \mathrm{k}^{2}+\mathrm{k}-6=0 \\ \Rightarrow \mathrm{k}=-3,2 \\ \mathrm{So}, \mathrm{k}=2(\mathrm{k}>0)

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himanshu.meshram

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