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JEE Main & Advanced Mathematics Conic Sections Question Bank Parabola

  • question_answer
    If \[{{x}^{2}}+6x+20y-51=0\], then axis of parabola is  [Orissa JEE 2004]

    A)            \[x+3=0\]                                  

    B)            \[x-3=0\]

    C)            \[x=1\]                                      

    D)            \[x+1=0\]

    Correct Answer: A

    Solution :

               Given equation of parabola is \[{{x}^{2}}+6x+20y-51=0\]                    \[\Rightarrow \]\[{{x}^{2}}+6x=-20y+51\]                    \[\Rightarrow \]\[{{(x+3)}^{2}}=-20y+60\Rightarrow {{(x+3)}^{2}}=-20(y-3)\]                    \[\Rightarrow \]\[{{(x+3)}^{2}}=-4.5(y-3)\]                    \[\therefore \] Axis of parabola is \[x+3=0\].


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