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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-II

  • question_answer
    Tangents drawn from the point \[(-8,0)\] to the parabola \[{{y}^{2}}=8x\] touch the parabola at \[P\] and \[Q\]. If \[F\] is the focus of the parabola, then the area of the triangle \[PFQ\] (in sq. units) is equal to [JEE Online 15-04-2018 (II)]

    A)         48                                          

    B) 32          

    C) 24                          

    D)          64

    Correct Answer: A

    Solution :

    Equation of the chord of contact PQ is given by \[T=0\] \[T\equiv 4(x+{{x}_{1}})-y{{y}_{1}}=0,\] where \[({{x}_{1}},{{y}_{1}})\equiv (-8,0)\] \[\therefore \] chord of contact is \[x=8\] Coordinates of point P and Q are \[(8,8)\] and \[(8,-8)\] Focus of the parabola is \[F(2,0)\] Area of triangle \[PQF=\frac{1}{2}\times (8-2)\times (8+8)=48sq.\text{units}\]


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