The x-intercept of the tangent to a curve is | vedclass

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(A) Let the point of contact be $(x, y)$. The equation of the tangent at $(x, y)$ is $Y - y = \frac{dy}{dx}(X - x)$.For the $x-$intercept,set $Y = 0$:

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$y\,e^{\frac{x}{y}} = e$

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(A) Let the point of contact be $(x, y)$. The equation of the tangent at $(x, y)$ is $Y - y = \frac{dy}{dx}(X - x)$.For the $x-$intercept,set $Y = 0$:$-y = \frac{dy}{dx}(X - x) \Rightarrow X - x = -\frac{y}{dy/dx} \Rightarrow X = x - y\frac{dx}{dy}$.Given that the $x-$intercept is equal to the ordinate $y$,we have $x - y\frac{dx}{dy} = y$.Rearranging gives $y\frac{dx}{dy} = x - y$,or $\frac{dx}{dy} = \frac{x}{y} - 1$.This is a linear differential equation in $x$: $\frac{dx}{dy} - \frac{1}{y}x = -1$.The integrating factor is $IF = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.Multiplying by $IF$: $\frac{1}{y} \frac{dx}{dy} - \frac{1}{y^2} x = -\frac{1}{y}$.Integrating both sides with respect to $y$: $\frac{x}{y} = -\ln|y| + C$.Since the curve passes through $(1, 1)$,we have $\frac{1}{1} = -\ln(1) + C \Rightarrow C = 1$.Thus,$\frac{x}{y} = 1 - \ln y \Rightarrow \ln y = 1 - \frac{x}{y}$.Taking the exponential of both sides: $y = e^{1 - \frac{x}{y}} = e \cdot e^{-\frac{x}{y}}$.Multiplying by $e^{\frac{x}{y}}$ gives $y\,e^{\frac{x}{y}} = e$.

The $x-$intercept of the tangent to a curve is equal to the ordinate of the point of contact. The equation of the curve passing through the point $(1, 1)$ is

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