Let $\mathrm{u}=\frac{2 \mathrm{z}+\mathrm{i}}{\mathrm{z}-\mathrm{ki}}, \mathrm{z}=\mathrm{x}+\mathrm{iy}$ and $\mathrm{k}>0$. If the curve represented by $\operatorname{Re}(\mathrm{u})+\operatorname{Im}(\mathrm{u})=1$ intersects the $y$-axis at the points $P$ and $Q$ where $PQ=5$, then the value of $k$ is:

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Complex Numbers

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:

Answer: (A)

Put

z = x + i y

u = \frac{2 z + i}{z - k i}

and separate real and imaginary parts; the condition

Re (u) + Im (u) = 1

reduces to

x^{2} + y^{2} + y - k - k^{2} = 0

. On the

y

-axis

x = 0

, so

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

, whose roots

y_{1}, y_{2}

satisfy

y_{1} + y_{2} = - 1

and

y_{1} y_{2} = - (k + k^{2})

. Given

P Q = ∣ y_{1} - y_{2} ∣ = 5

(y_{1} - y_{2})^{2} = (y_{1} + y_{2})^{2} - 4 y_{1} y_{2} = 1 + 4 (k + k^{2}) = 25

, so

k^{2} + k - 6 = 0

, giving

k = 2

k = - 3

. Since

k > 0

k = 2

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option