What is the equation of the normal to the cur | vedclass

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(A) The given line is $2x - 2y + 3 = 0$,which can be written as $y = x + \frac{3}{2}$. The slope of this line is $m = 1$.Since the normal is parallel

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$x - y = 3e^{-2}$

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(A) The given line is $2x - 2y + 3 = 0$,which can be written as $y = x + \frac{3}{2}$. The slope of this line is $m = 1$.Since the normal is parallel to this line,the slope of the normal is $m_n = 1$.The slope of the tangent to the curve $y = x \log x$ is $\frac{dy}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1$.The slope of the normal is $m_n = -\frac{1}{dy/dx} = -\frac{1}{\log x + 1}$.Equating the slopes: $-\frac{1}{\log x + 1} = 1 \implies \log x + 1 = -1 \implies \log x = -2 \implies x = e^{-2}$.For $x = e^{-2}$,the value of $y$ is $y = e^{-2} \log(e^{-2}) = e^{-2} (-2) = -2e^{-2}$.The point of contact is $(e^{-2}, -2e^{-2})$.The equation of the normal is $y - y_1 = m_n(x - x_1)$.$y - (-2e^{-2}) = 1(x - e^{-2})$.$y + 2e^{-2} = x - e^{-2}$.$x - y = 3e^{-2}$.

What is the equation of the normal to the curve $y = x \log x$ which is parallel to the line $2x - 2y + 3 = 0$?

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