JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Sphere

  • question_answer
    A point moves so that the sum of its distances from the points (4, 0, 0) and (?4, 0, 0) remains 10. The locus of the point is         [MP PET 1988]

    A)            \[9{{x}^{2}}-25{{y}^{2}}+25{{z}^{2}}=225\]                                 

    B)            \[9{{x}^{2}}+25{{y}^{2}}-25{{z}^{2}}=225\]

    C)            \[9{{x}^{2}}+25{{y}^{2}}+25{{z}^{2}}=225\]                                 

    D)            \[9{{x}^{2}}+25{{y}^{2}}+25{{z}^{2}}+225=0\]

    Correct Answer: C

    Solution :

                       \[\sqrt{{{(x-4)}^{2}}+{{y}^{2}}+{{z}^{2}}}+\sqrt{{{(x+4)}^{2}}+{{y}^{2}}+{{z}^{2}}}=10\]            \[\Rightarrow \,\,2\,({{x}^{2}}+{{y}^{2}}+{{z}^{2}})+2\sqrt{[{{(x-4)}^{2}}+{{y}^{2}}+{{z}^{2}}]\,[{{(x+4)}^{2}}+{{y}^{2}}+{{z}^{2}}]}\]                    \[=100-32=68\]            \[\Rightarrow \,\,{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-34)}^{2}}\]            \[=[{{(x-4)}^{2}}+{{y}^{2}}+{{z}^{2}}]\,[{{(x+4)}^{2}}+{{y}^{2}}+{{z}^{2}}]\]            \[\Rightarrow \,\,{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{2}}-68\,({{x}^{2}}+{{y}^{2}}+{{z}^{2}})+{{(34)}^{2}}\]            \[=[({{x}^{2}}+{{y}^{2}}+{{z}^{2}}+16)-8x]\,\,[({{x}^{2}}+{{y}^{2}}+{{z}^{2}}+16)+8x]\]            \[={{({{x}^{2}}+{{y}^{2}}+{{z}^{2}}+16)}^{2}}-64{{x}^{2}}\]            \[=\,({{x}^{2}}+{{y}^{2}}+{{z}^{2}})+32\,({{x}^{2}}+{{y}^{2}}+{{z}^{2}})-64{{x}^{2}}+{{(16)}^{2}}\]                    \[\Rightarrow \,\,9{{x}^{2}}+25{{y}^{2}}+25{{z}^{2}}-225=0\].


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