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Question: Let \(z = 1 - t + i\sqrt {{t^2} + t + 2} \) where t is a real parameter. The locus of Z in the argan...

Let z=1−t+it2+t+2z = 1 - t + i\sqrt {{t^2} + t + 2} where t is a real parameter. The locus of Z in the argand plane is
A. a hyperbola  B. an ellipse  C. a straight line  D. none of these  {\text{A}}{\text{. a hyperbola }} \\\ {\text{B}}{\text{. an ellipse }} \\\ {\text{C}}{\text{. a straight line }} \\\ {\text{D}}{\text{. none of these}} \\\

Explanation

Solution

Hint: We have to compare with x+iyx + iy and solve the equations to remove t and get an answer in x and y. And here the argand plane means the plane in which we represent a complex number.

Complete step-by-step answer:
We have
z=1−t+it2+t+2z = 1 - t + i\sqrt {{t^2} + t + 2}
Here we compare with general form of complex equation
z=x+iyz = x + iy
∴x=1−t,y=t2+t+2\therefore x = 1 - t,y = \sqrt {{t^2} + t + 2}
Now we have to eliminate t and get the equation in x and y so that we can check which option is the correct option.
⇒t=1−x\Rightarrow t = 1 - x and also y2=t2+t+2{y^2} = {t^2} + t + 2
So from here we will eliminate t by putting t=1-x
∴y2=(1−x)2+1−x+2\therefore {y^2} = {\left( {1 - x} \right)^2} + 1 - x + 2
⇒y2=1+x2+−2x+1−x+2\Rightarrow {y^2} = 1 + {x^2} + - 2x + 1 - x + 2
⇒y2=x2−3x+4\Rightarrow {y^2} = {x^2} - 3x + 4
⇒y2−x2+3x−4=0\Rightarrow {y^2} - {x^2} + 3x - 4 = 0
Here we compare with general equation of second degree ax2+2hxy+by2+2gx+2fy+c=0a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0
We get a= -1, b =1 , h = 0
So h2>ab{h^2} > ab ( condition of hyperbola)
So this is the equation of a hyperbola.
Option A is the correct option.

Note: Whenever you get this type of question the key concept of solving is first we have to consider the general form of complex number and then on comparing it from question and on further solving we will get an equation in x and y and we have to just check through the condition of conic.