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

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(B) Given curve is $y=x \log x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$. The s

Vedclass · Accepted Answer

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

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(B) Given curve is $y=x \log x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$. The slope of the tangent at any point $(x, y)$ is $m_t = 1 + \log x$. The slope of the normal $m_n$ is given by $m_n = -\frac{1}{m_t} = -\frac{1}{1 + \log x}$. The given line is $2x - 2y + 3 = 0$,which can be written as $y = x + \frac{3}{2}$. The slope of this line is $1$. Since the normal is parallel to the given line,its slope must be equal to $1$. Therefore,$-\frac{1}{1 + \log x} = 1 \implies 1 + \log x = -1 \implies \log x = -2 \implies x = e^{-2}$. Substituting $x = e^{-2}$ in the curve equation,$y = e^{-2} \log(e^{-2}) = e^{-2}(-2) = -2e^{-2}$. The point of contact is $(e^{-2}, -2e^{-2})$. The equation of the normal with slope $1$ passing through $(e^{-2}, -2e^{-2})$ is $y - (-2e^{-2}) = 1(x - e^{-2})$. $y + 2e^{-2} = x - e^{-2} \implies x - y = 3e^{-2}$.

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

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