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JCECE Engineering JCECE Engineering Solved Paper-2002

  • question_answer
    The centre of the ellipse\[\frac{{{(x+y-2)}^{2}}}{9}+\frac{{{(x-y)}^{2}}}{16}=1\]is:

    A) \[(0,\,\,0)\]                        

    B) \[(1,\,\,1)\]

    C)  \[(1,\,\,0)\]                       

    D)  \[(0,\,\,1)\]

    Correct Answer: B

    Solution :

    Key Idea: If the equation of ellipse be\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], then centre of ellipse be\[(0,\,\,0)\]. Given equation of ellipse be                 \[\frac{{{(x+y-2)}^{2}}}{9}+\frac{{{(x-y)}^{2}}}{16}=1\] For finding a centre, we put                 \[x+y-2=0\]and\[x-y=0\] \[\Rightarrow \]               \[y=1,\,\,x=1\] \[\therefore \]Required centre is\[(1,\,\,1)\].


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