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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    If the area  of the  triangle  whose one vertex is at the vertex of the parabola, \[{{y}^{2}}+4(x-{{a}^{2}})=0\] and the other two vertices are the points of intersection of the parabola and y-axis. is 250 sq. units, then a value of a is [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) \[5\sqrt{5}\]                                          

    B) \[{{(10)}^{2/3}}\]

    C) \[5{{(2)}^{1/3}}\]                   

    D)   5

    Correct Answer: D

    Solution :

    The vertex of the given parabola is\[({{a}^{2}},0).\] When \[x=0,{{y}^{2}}+4(0-{{a}^{2}})=0\Rightarrow y=\pm 2a\] The point of intersection of the given parabola and the y-axis are \[\left( 0,\pm 2a \right).\] \[\therefore \]Area of the triangle \[=\frac{1}{2}.4a.{{a}^{2}}\] \[\Rightarrow \]\[250=2{{a}^{3}}\Rightarrow a=5\]


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