JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 10)
Let $$u = {{2z + i} \over {z - ki}}$$, 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 PQ = 5, then the value of k is :
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 :
2
4
1/2
3/2
Explanation
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)
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)
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