JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer The equation of a common tangent to \[{{y}^{2}}=4x\] and the curve \[{{x}^{2}}+4{{y}^{2}}=8\]can be

    A) \[x-2y+2=0\] 

    B) \[x+2y+4=0\]

    C) \[x-2y=4\]                   

    D) \[x+2y=4\]

    Correct Answer: B

    Solution :

    \[{{y}^{2}}=4x\And \frac{{{x}^{2}}}{8}+\frac{{{y}^{2}}}{2}=1\] Equation of tangent to above curves are respectively. \[{{y}^{2}}=mx+\frac{1}{m}\]and\[y=mx+\sqrt{8{{m}^{2}}+2}\] Comparing\[\frac{1}{m}=\sqrt{8{{m}^{2}}+2}\]\[\Rightarrow \]\[{{m}^{2}}(8{{m}^{2}}+2)=1\]seeing the options \[m=\pm \frac{1}{2}\]satisfy the equation \[\Rightarrow \]\[y=\pm \frac{1}{2}x\pm 2\Rightarrow 2y=\pm x\pm 4\] \[i.e.\,\,2y=x+4\And x+2y+4=0\]


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