P is the point of contact of the tangent draw | vedclass

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(D) Given the curve $y = \log_{e} x$ $(i)$. Let the point of contact be $P(\alpha, \beta)$. The slope of the tangent is $\frac{dy}{dx} = \frac{1}{x}$.

Vedclass · Accepted Answer

$\sqrt{e^{2}+1}$

Vedclass · Answer

(D) Given the curve $y = \log_{e} x$ $(i)$. Let the point of contact be $P(\alpha, \beta)$. The slope of the tangent is $\frac{dy}{dx} = \frac{1}{x}$. The equation of the tangent at $P$ is $(y - \beta) = \frac{1}{\alpha}(x - \alpha)$. Since the tangent passes through the origin $(0,0)$,we have $(0 - \beta) = \frac{1}{\alpha}(0 - \alpha)$,which gives $\beta = 1$. From equation $(i)$,at $P$,$\beta = \log_{e} \alpha$. Since $\beta = 1$,we have $1 = \log_{e} \alpha$,so $\alpha = e$. Thus,the point of contact is $P(e, 1)$. The slope of the tangent at $P$ is $\frac{1}{e}$,so the slope of the normal at $P$ is $-e$. The equation of the normal at $P(e, 1)$ is $(y - 1) = -e(x - e)$,which simplifies to $ex + y - (e^{2} + 1) = 0$. The length of the perpendicular from the origin $(0,0)$ to the line $ex + y - (e^{2} + 1) = 0$ is given by $d = \frac{|e(0) + 1(0) - (e^{2} + 1)|}{\sqrt{e^{2} + 1^{2}}} = \frac{e^{2} + 1}{\sqrt{e^{2} + 1}} = \sqrt{e^{2} + 1}$.

$P$ is the point of contact of the tangent drawn from the origin to the curve $y = \log_{e} x$. The length of the perpendicular drawn from the origin to the normal at $P$ is

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