A) an ellipse
B) hyperbola
C) a straight line
D) None of these
Correct Answer: B
Solution :
Let \[\operatorname{z}=\,\,1+t+i\sqrt{{{t}^{2}}+t\,+2}\] We know that \[\operatorname{z}=\,\,x\,+ iy\] \[\operatorname{x}\,+\,iy=\,\,1-t\,\,+\,\,i\,\sqrt{{{t}^{2}}+t+2}\] compare real and imaginary part, we get \[\operatorname{x}=1-t\,\,\Rightarrow \,\,t=1-x\] and \[\operatorname{y}=\,\,\sqrt{{{t}^{2}}+t\,+2}\] \[\Rightarrow \,\,\,\,{{y}^{2}}={{t}^{2}}+t+2\] \[\Rightarrow \,\,\,{{y}^{2}}=\,\,{{(1-x)}^{2}}+\,\,(1-x)+2\] \[{{\operatorname{y}}^{2}}=\,\,1+{{x}^{2}}-2x+1-x+2\] \[\Rightarrow \,\,\,\,\,{{y}^{2}}={{x}^{2}}-3x+4\] \[\Rightarrow \,\,\,\,\,{{y}^{2}}={{\left( x-\frac{3}{2} \right)}^{2}}+\frac{7}{4}\] Which is a hyperbola.
You need to login to perform this action.
You will be redirected in
3 sec
