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Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    A parabola is drawn with its focus at\[(3,\,\,4)\]and vertex at the focus of the parabola\[{{y}^{2}}-12x-4y+4=0\]. The equation of the parabola is

    A) \[{{y}^{2}}-8x-6y+25=0\]

    B) \[{{y}^{2}}-6x+8y-25=0\]

    C) \[{{x}^{2}}-6x-8y+25=0\]

    D) \[{{x}^{2}}+6x-8y-25=0\]

    Correct Answer: C

    Solution :

    Given equation can be rewritten as                 \[{{(y-2)}^{2}}=12x\] Here, vertex and foci are (0, 2) and (3, 2). \[\therefore \]Vertex of the required parabola is (3, 2) and focus is (3, 4). The axis of symmetry is\[x=3\]and latusrectum\[=4\cdot 2=8\] Hence, required equation is                                 \[{{(x-3)}^{2}}=8(y-2)\] or                            \[{{x}^{2}}-6x-8y+25=0\]


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