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(C) Given curve is $y = x \log x$. Finding the derivative: $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. The slope of the tange

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$(\frac{1}{e^2}, \frac{-2}{e^2})$

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(C) Given curve is $y = x \log x$. Finding the derivative: $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. The slope of the tangent at point $P(x, y)$ is $m_t = \log x + 1$. The slope of the normal at point $P$ is $m_n = -\frac{1}{m_t} = -\frac{1}{\log x + 1}$. The given line is $2x - 2y = 3$,which can be written as $y = x - \frac{3}{2}$. The slope of this line is $m_l = 1$. Since the normal is parallel to the line,$m_n = m_l$,so $-\frac{1}{\log x + 1} = 1$. This implies $\log x + 1 = -1$,so $\log x = -2$. Thus,$x = e^{-2} = \frac{1}{e^2}$. Substituting $x$ into the curve equation: $y = \frac{1}{e^2} \log(\frac{1}{e^2}) = \frac{1}{e^2} (-2) = -\frac{2}{e^2}$. Therefore,the point $P$ is $(\frac{1}{e^2}, -\frac{2}{e^2})$.

If the normal drawn at the point $P$ on the curve $y=x \log x$ is parallel to the line $2x-2y=3$,then $P=$

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