JEE Main & Advanced Sample Paper JEE Main Sample Paper-45

  • question_answer
    The point of contact of a common tangent to the parabola \[{{y}^{2}}=8x\] and an ellipse \[15{{x}^{2}}+4{{y}^{2}}=60\] is

    A)  \[\left( \frac{1}{2},\,\frac{15}{4} \right)\]                 

    B)  \[\left( -\frac{1}{2},\,\frac{15}{4} \right)\]

    C)  \[\left( \frac{1}{2},-\,\frac{15}{4} \right)\]               

    D)  \[\left( -\frac{1}{2},-\,\frac{15}{2} \right)\]

    Correct Answer: B

    Solution :

     Let the point of contact with ellipse is \[({{x}_{1}},\,\,{{y}_{1}})\]. Then, \[15x{{x}_{1}}+4y{{y}_{1}}-60=0\]           ?(i) Let the tangent of \[{{y}^{2}}=8x\,\] is \[y=\,mx+\frac{2}{m}\] and the tangent of \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{15}=1\] is \[y=mx\pm \,\sqrt{4{{m}^{2}}+15}\] \[\therefore \]    \[\frac{2}{m}=\pm \,\sqrt{4{{m}^{2}}+15}\] \[\Rightarrow \]            \[4{{m}^{4}}+15{{m}^{2}}-4=0\] \[\Rightarrow \]            \[m=\pm \frac{1}{2}\] Hence, equation of common tangent at \[m=\frac{1}{2}\] \[\Rightarrow \]            \[y=\frac{x}{2}+4\] \[\Rightarrow \]            \[x-2y+8=0\]              ?(ii) From Eqs. (i) and  (ii), we get \[\frac{15{{x}_{1}}}{1}=\frac{4{{y}_{1}}}{-2}=\frac{-60}{8}\] \[\Rightarrow \]            \[{{x}_{1}}=-\frac{1}{2}\]  and      \[{{y}_{1}}=\frac{15}{4}\]

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