Co-ordinates of a point on the curve y = x lo | vedclass

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(D) Given the curve $y = x \log x$. First,find the derivative $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$. The slope of the ta

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$(e^{-2}, -2e^{-2})$

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(D) Given the curve $y = x \log x$. First,find the derivative $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$. The slope of the tangent at any point $(x, y)$ is $m_t = 1 + \log x$. The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{1 + \log x}$. The given line is $2x - 2y = 3$,which can be written as $y = x - \frac{3}{2}$. The slope of this line is $1$. Since the normal is parallel to the line,their slopes must be equal: $-\frac{1}{1 + \log x} = 1$ $-1 = 1 + \log x$ $\log x = -2$ $x = e^{-2}$. Now,substitute $x = e^{-2}$ into the curve equation: $y = e^{-2} \log(e^{-2}) = e^{-2} (-2) = -2e^{-2}$. Thus,the required point is $(e^{-2}, -2e^{-2})$.

Co-ordinates of a point on the curve $y = x \log x$ at which the normal is parallel to the line $2x - 2y = 3$ are

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