RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    The region of the argand diagram defined by\[|z-1|+|z+1|\le 4\]is

    A)  interior of an ellipse

    B)  exterior of a circle

    C)  interior and boundary of an ellipse

    D)  None of the above

    Correct Answer: C

    Solution :

    Let P,A and B represent complex number \[z,1+0i,-1+0i\]respectively, then \[|z-1|+|z+1|\le 4\] \[\Rightarrow \] \[PA+PB\le 4\] \[\Rightarrow \]P moves in such a way that the sum of its distance from two fixed points is always less than or equal to 4 (which is greater than distance between). \[\Rightarrow \]Locus of P is the interior and boundary of ellipse having foci at (1, 0) and (-1, 0). Aliter Let              \[z=x+iy\] Given,     \[|z-1|+|z+1|\le 4\] \[\Rightarrow \] \[|(x-1)+iy|+(x+1)+iy|\le 4\] \[\Rightarrow \]\[\sqrt{{{(x-1)}^{2}}+{{y}^{2}}}+\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}\le 4\] \[\Rightarrow \]\[{{(x-1)}^{2}}+{{y}^{2}}\le 16+{{(x+1)}^{2}}+{{y}^{2}}\] \[-8\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}\] \[\Rightarrow \] \[8\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}\le 4x+16\] \[\Rightarrow \] \[2\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}\le x+4\] \[\Rightarrow \] \[4({{x}^{2}}+{{y}^{2}}+2x+1)\le {{x}^{2}}+8x+16\] \[\Rightarrow \] \[3{{x}^{2}}+4{{y}^{2}}\le 12\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}\le 1\] which represents boundary and interior of an ellipse \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=1\]


You need to login to perform this action.
You will be redirected in 3 sec spinner