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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    A particle is moving with a velocity \[\overrightarrow{v}=K\left( y\,\,\overrightarrow{i}+x\,\,\overrightarrow{j} \right)\], where K is a constant. The general equation for its path is: [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) \[y={{x}^{2}}+constant\]

    B) \[{{y}^{2}}=\text{ }{{x}^{2}}+constant\]

    C) \[{{y}^{2}}=\,\,x+constant\]

    D) \[xy\text{ }=\text{ }constant\]

    Correct Answer: B

    Solution :

    \[\frac{dx}{dt}\,\,=\,\,y\] \[\frac{dy}{dt}\,\,=\,\,x\] \[\frac{dx}{dy}\,\,=\,\,\frac{y}{x}\] \[xdx\text{ }=\text{ }ydy\] \[{{y}^{2}}={{x}^{2}}+C\]


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