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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    The directrix of the parabola\[{{x}^{2}}-4x-8y+12=0\] is :

    A)  \[y=0\]                               

    B)  \[x=1\]

    C)  \[y=-1\]              

    D)  \[x=-1\]

    Correct Answer: C

    Solution :

    Here we have                 \[{{x}^{2}}-4x-8y+12=0\] \[\Rightarrow \]               \[{{x}^{2}}-4x+4-8y+8=0\] \[\Rightarrow \]               \[{{(x-2)}^{2}}-8y+8=0\] \[\Rightarrow \]                               \[{{(x-2)}^{2}}=8\,(y-1)\] So this is the form \[{{X}^{2}}=4aY\] where \[X=x-2,\,Y=y-1\] and \[a=2\] So directrix \[x\] is given by                 \[Y=-a\]                 \[y-1=-2\]                 \[y=-2+1\]                 \[y=-1\]


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