JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    Let P be the point of intersection of the common tangents to the parabola \[{{y}^{2}}=12x\] and the hyperbola \[8{{x}^{2}}{{y}^{2}}=8.\]If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio:                                                                [JEE Main Held on 12-4-2019 Morning]

    A) 5:4      

    B) 14:13

    C) 2:1                              

    D) 13:11

    Correct Answer: A

    Solution :

    Equation of tangents \[{{y}^{2}}=12x\Rightarrow y=2x+\frac{3}{m}\] \[\frac{{{x}^{2}}}{1}-\frac{{{y}^{2}}}{8}=1\Rightarrow y=mx\pm \sqrt{{{m}^{2}}-8}\] Since they are common tangent \[\therefore \]\[\frac{3}{m}=\pm \sqrt{{{m}^{2}}-8}\]      \[\frac{{{x}^{2}}}{1}-\frac{{{y}^{2}}}{8}=1\] \[{{m}^{4}}-8{{m}^{2}}-9=0\]  \[e=3\] \[m=\pm 3\]                            \[ae=3\]              


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