JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    Let\[z\in C,\]the set of complex numbers. Then the equation, \[2|z+3i|-|z-i|=0\]a circle with radius\[\frac{8}{3}.\]                [JEE Online 08-04-2017]

    A)  a circle with radius\[\frac{8}{3}\]

    B)  an ellipse with length of minor axis\[\frac{16}{9}\]

    C)  an ellipse with length of major axis\[\frac{16}{3}\]

    D)  a circle with diameter\[\frac{10}{3}\]

    Correct Answer: A

    Solution :

    \[2|x+i(y+3)|=|x+i(y-1)|\] \[=2\sqrt{{{x}^{2}}+{{(y+3)}^{2}}}\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}\] \[=4{{x}^{2}}+4{{(y+3)}^{2}}={{x}^{2}}+{{(y-1)}^{2}}\] \[=3{{x}^{2}}={{y}^{2}}-2y+1-4{{y}^{2}}-24y-36\] \[=3{{x}^{2}}+3{{y}^{2}}+26y+35=0\] \[={{x}^{2}}+{{y}^{2}}+\frac{26}{3}y+\frac{35}{3}=0\] \[=r=\sqrt{0+\frac{169}{9}-\frac{35}{3}}\] \[=\sqrt{\frac{64}{9}}=\frac{8}{3}\]         


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