Let u =(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 :

Question

Let u =(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 : (A) 3 / 2 (B) 4 (C) 2 (D) 1 / 2

Tardigrade Admin · Accepted Answer

u =(2 z + i / z - ki ) =(2 x 2+(2 y +1)( y - k )/ x 2+( y - k )2)+ i (( x (2 y +1)-2 x ( y - k ))/ x 2+( y - k )2) Since Re( u )+Im( u )=1 ⇒ 2 x 2+(2 y +1)( y - k )+ x (2 y +1)-2 x ( y - k ) =x2+( y - k )2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/73b50b6484ad8dcfe631ee134e17a732-.png" /> ∵ PQ = 5 ⇒ mid y 1- y 2 | =5 ⇒ k 2+ k -6=0 ⇒ k =-3,2 So , k =2( k >0)

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

$3/2$

$4$

$2$

$1/2$

Q. Let u=z−ki2z+i​,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 PQ=5, then the value of k is :

3/2

4

2

1/2

u=z−ki2z+i​ =x2+(y−k)22x2+(2y+1)(y−k)​+ix2+(y−k)2(x(2y+1)−2x(y−k))​ Since Re(u)+Im(u)=1 ⇒2x2+(2y+1)(y−k)+x(2y+1)−2x(y−k) =x2+(y−k)2 ∵PQ=5 ⇒∣y1​−y2​∣=5 ⇒k2+k−6=0 ⇒k=−3,2 So,k=2(k>0)

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

$3/2$

$4$

$2$

$1/2$