11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • question_answer The equation of the common tangent to the curves \[{{y}^{2}}=8x\] and \[xy=-1\] is        [IIT Screening 2002]

    A)            \[3y=9x+2\]                               

    B)            \[y=2x+1\]

    C)            \[2y=x+8\]                                 

    D)            \[y=x+2\]

    Correct Answer: D

    Solution :

               Any point on \[{{y}^{2}}=8x\] is \[(2{{t}^{2}},\,4t)\] where the tangent is \[yt=x+2{{t}^{2}}.\]            Solving it with \[xy=-1,\] \[y(yt-2{{t}^{2}})=-1\]             or \[t{{y}^{2}}-2{{t}^{2}}y+1=0\].            For common tangent, it should have equal roots.                   \ \[4{{t}^{4}}-4t=0\] Þ \[t=0,\,1\].                      \[\therefore \] The common tangent is \[y=x+2\], (when \[t=0\], it is \[x=0\] which can touch \[xy=-1\] at infinity only).

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