The equation of the normal to the curve y=x l | vedclass

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(B) Given curve is $y=x \log x$ ... $(i)$ Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 + \log x$. The slope of the normal is given by

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

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(B) Given curve is $y=x \log x$ ... $(i)$ Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 + \log x$. The slope of the normal is given by $m_n = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{1+\log x}$. 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 the given line,their slopes must be equal: $-\frac{1}{1+\log x} = 1$ $\Rightarrow 1+\log x = -1$ $\Rightarrow \log x = -2$ $\Rightarrow x = e^{-2}$. Substituting $x = e^{-2}$ in $(i)$,we get $y = e^{-2} \log(e^{-2}) = e^{-2}(-2) = -2e^{-2}$. The point of contact is $(e^{-2}, -2e^{-2})$. The equation of the normal is $y - y_1 = m(x - x_1)$: $y - (-2e^{-2}) = 1(x - e^{-2})$ $y + 2e^{-2} = x - e^{-2}$ $x - y = 3e^{-2}$.

The equation of the normal to the curve $y=x \log x$ parallel to $2x-2y+3=0$ is

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