BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    The equation of the hyperbola whose directrix is \[x+2y=1,\]focus \[(2,1)\] and eccentricity 2 will be:

    A)  \[{{x}^{2}}-16xy-11{{y}^{2}}-12x+6y+21=0\]

    B)  \[3{{x}^{2}}+16xy+15{{y}^{2}}-4x-14y+1=0\]

    C)  \[{{x}^{2}}+16xy+11{{y}^{2}}-12x+6y+21=0\]

    D)  none of the above

    Correct Answer: A

    Solution :

    Key Idea: If P be any point on the hyperbola, S be a focus and M be a point on the directrix, then \[PS=ePM.\] Since, \[S=(2,1),e=2\] Equation of directrix is \[x+2y=1,\]                 \[\because \]     \[PS=ePM\]                 \[\therefore \]  \[{{(x-2)}^{2}}+{{(y-1)}^{2}}=4{{\left( \frac{x+2y-1}{\sqrt{{{1}^{2}}+{{2}^{2}}}} \right)}^{2}}\]                 \[\Rightarrow \]               \[5({{x}^{2}}+{{y}^{2}}-4x-2y+5)\]            \[\Rightarrow \]               \[=4({{x}^{2}}+4{{y}^{2}}+1+4xy-4y-2x)\] \[\Rightarrow \]               \[{{x}^{2}}-11{{y}^{2}}-16xy-12x+6y+21=0\]


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