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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2005

  • question_answer
    The ends of the latus rectum of the conic\[{{x}^{2}}+10x-16y+25=0\] are :

    A)  (3,-4), (13, 4)

    B)  (-3,-4), (13,-4)

    C)  (3, 4), (-13, 4)

    D)  (5,-8), (-5, 8)

    Correct Answer: C

    Solution :

    Given that                 \[{{x}^{2}}+10x-16y+25=0\] \[\Rightarrow \,\,{{(x+5)}^{2}}=16\,y\] \[\Rightarrow \]               \[{{X}^{2}}=4\,AY\] where \[X=x+5,\,A=4,\,Y=y\]. The ends of die latus rectum are \[\left( +2A,A \right)\] and \[\left( -2A,A \right)\] \[\Rightarrow \]               \[x+5=2\,(4)\Rightarrow x=-8-5=3,\,y=4\] and \[x+5=-2\,(4)\Rightarrow x=-8-5=-13,\,y=4\] \[\Rightarrow \]               (3, 4) and (-13, 4).


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