Question

Let $z = 1 - t + i\sqrt {{t^2} + t + 2}$ 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

Answer 1

Hint: Assume a general complex number and then compare it with the number given, this will lead you to different numbers both will be real and use them to find the locus of z.
Complete Step by Step Solution:
Let us assume a complex number $z = x + iy$ . If we try to compare this equation with the complex number given, i,e,, the real part with the real part and the complex part with the complex. Then we can get the values of x and y as $1 - t$ and $\sqrt {{t^2} + t + 2}$ respectively now for x we can rewrite it as

\therefore x = 1 - t\\\ \Rightarrow t = 1 - x.............................................(i) \end{array}$$ And for y we can rewrite it as $$\begin{array}{l} \therefore y = \sqrt {{t^2} + t + 2} \\\ \Rightarrow {y^2} = {t^2} + t + 2......................................(ii) \end{array}$$ Now putting the value of t from equation (i) in equation (ii) We will get it as $$\begin{array}{*{20}{l}} {\therefore {y^2} = {{(1 - x)}^2} + (1 - x) + 2}\\\ { \Rightarrow {y^2} = {{(1 - x)}^2} + 3 - x}\\\ { \Rightarrow {y^2} = 1 + {x^2} - 2x + 3 - x}\\\ { \Rightarrow {y^2} = {x^2} - 3x + 4}\\\ \begin{array}{l} \Rightarrow {y^2} - {x^2} + 3x - 4 = 0\\\ \Rightarrow {y^2} - \left( {{x^2} - 2 \times \dfrac{3}{2} \times x + {{\left( {\dfrac{3}{2}} \right)}^2}} \right) = 4 + {\left( {\dfrac{3}{2}} \right)^2}\\\ \Rightarrow {y^2} - {\left( {x - \dfrac{3}{2}} \right)^2} = 4 + \dfrac{9}{4}\\\ \Rightarrow {y^2} - {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{25}}{4}\\\ \Rightarrow \dfrac{{4{y^2}}}{{25}} - \dfrac{{4{{\left( {x - \dfrac{3}{2}} \right)}^2}}}{{25}} = 1\\\ \Rightarrow \dfrac{{{{\left( {2y} \right)}^2}}}{{25}} - \dfrac{{{{\left( {2x - 3} \right)}^2}}}{{25}} = 1 \end{array} \end{array}$$ So if we take $$Y = 2y\& X = 2x - 3$$ Now it is clear that it is not a straight line therefore we have to check for hyperbola and ellipse The general equation of hyperbola is given by $$\dfrac{{{X^2}}}{{{A^2}}} - \dfrac{{{Y^2}}}{{{B^2}}} = 1$$ So it is clear that the given equation is a hyperbola And option A is correct. Note: While getting the equation I have used the algebraic formula $${(a + b)^2} = {a^2} + {b^2} + 2ab$$ . After getting the equation keep in mind the squared terms are in addition for an ellipse and in subtraction for an hyperbola as in this case it was a negative sign in between the squared variables therefore it is concluded as an hyperbola.