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

  • question_answer
    Direction: A straight line will touch a given conic if there is only one point of intersection of the line and the given conic. If the conic is specified by quadratic equation in\[x\] and \[y,\] then the straight line will touch if the discriminant of the equation obtained by the elimination of one of the variable is zero. Let us consider parabola \[{{y}^{2}}=8x\] and an ellipse\[15{{x}^{2}}+4{{y}^{2}}=60\].
    The equation of a tangent common to both the parabola and the ellipse is

    A)  \[2x-y-8=0\]  

    B)  \[x-2y+8=0\]

    C)  \[2x-y+8=0\]

    D)  \[x+2y-8=0\]

    Correct Answer: B

    Solution :

     Let the tangent of \[{{y}^{2}}=8x\] is \[y=mx+2\,/m\] and the tangent of \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{15}=1\] is \[y=mx\,\pm \,\sqrt{4{{m}^{2}}+15}\] From Eqs. (i) and (ii), we get \[\frac{2}{m}=\pm \,\sqrt{4{{m}^{2}}+15}\] \[\Rightarrow \]            \[4{{m}^{4}}+15{{m}^{2}}-4=0\] \[\Rightarrow \]            \[m=\pm \,\frac{1}{2}\] Hence, equation of tangent is \[atm=\frac{1}{2}\] \[\Rightarrow \]            \[y=\frac{x}{2}+4\] \[\Rightarrow \]            \[2y=x+8\]

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