question_answer

Passage (Q. - 63) If z satisfies \[|z-(2-\sqrt{3}+i)|+|z-(2+\sqrt{3}+i)|=4.\]If a and b are two complex numbers which satisfies above equations which corresponds to maximum & minimum value of arg (z) respectively. If there exist a complex number 'c' which satisfy the equation, such that area of triangle formed by a, b, c is maximum. Then Complex number C is given by

A) \[(2+\sqrt{2})+i\left( 1-\frac{1}{\sqrt{2}} \right)\]

B) \[(2+\sqrt{2})+i\left( 1+\frac{1}{\sqrt{2}} \right)\]

C) \[(2-\sqrt{2})+i\left( 1-\frac{1}{\sqrt{2}} \right)\]

D) \[(2-\sqrt{2})+i\left( 1+\frac{1}{\sqrt{2}} \right)\]

Correct Answer: B

Solution :

\[|z-(2-\sqrt{3}+i)|+|z-(2+\sqrt{3}+i)|=4\]represents ellipse (as \[(as|{{z}_{1}}-{{z}_{2}}|<4)\] with foci as \[(2-\sqrt{3},1)\And (2+\sqrt{3},1)\]and length of major axis is 4\[\Rightarrow 2ae=2\sqrt{3}\Rightarrow e=\frac{\sqrt{3}}{2}\And b=1\] \[\therefore \]From figure the point with maximum & minimum argument are A & B. \[\therefore \]\[a=i,b=2\] Equation of ellipse is\[\frac{{{(x-2)}^{2}}}{4}+\frac{{{(y-1)}^{2}}}{1}=1\] For area of triangle ABC to be maximum, the tangent at C is parallel to AB. \[\Rightarrow \]slope of tangent at \[C=\frac{2}{\sin \frac{\theta }{1}}=-\frac{1}{2}\Rightarrow \theta =\frac{\pi }{4}\] \[C(2+2cos{{45}^{o}},1+1.sin{{45}^{o}})\] \[\equiv C\left( 2+\sqrt{2},1+\frac{1}{\sqrt{2}} \right)\]                                 (3)