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

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(A) Let the point of contact be $P(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.The equation of the tangent at $P(x, y)$ is $Y - y = \frac{dy}{

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

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(A) Let the point of contact be $P(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.The equation of the tangent at $P(x, y)$ is $Y - y = \frac{dy}{dx}(X - x)$.To find the $x-$intercept,set $Y = 0$: $-y = \frac{dy}{dx}(X - x) \Rightarrow X - x = -y \frac{dx}{dy} \Rightarrow X = x - y \frac{dx}{dy}$.According to the problem,the $x-$intercept is equal to the ordinate $y$,so $x - y \frac{dx}{dy} = y$.Rearranging the terms: $x - y = y \frac{dx}{dy} \Rightarrow (x - y) dy = y dx \Rightarrow x dy - y dy = y dx$.This can be written as $x dy - y dx = y dy$.Dividing both sides by $y^2$: $\frac{x dy - y dx}{y^2} = \frac{y dy}{y^2} = \frac{dy}{y}$.This is the derivative of the quotient: $d(\frac{x}{y}) = \frac{dy}{y}$.Integrating both sides: $\frac{x}{y} = \ln|y| + C$.Since the curve passes through $(1, 1)$,we substitute $x=1, y=1$: $\frac{1}{1} = \ln(1) + C \Rightarrow 1 = 0 + C \Rightarrow C = 1$.Thus,the equation is $\frac{x}{y} = \ln y + 1 \Rightarrow \frac{x}{y} - 1 = \ln y \Rightarrow \ln y = \frac{x-y}{y}$.Exponentiating both sides: $y = e^{\frac{x-y}{y}} = e^{\frac{x}{y} - 1} = e^{\frac{x}{y}} \cdot e^{-1}$.Therefore,$y e = e^{\frac{x}{y}} \Rightarrow y e^{1 - \frac{x}{y}} = \dots$ wait,let's re-evaluate: $y = e^{\frac{x}{y}} \cdot e^{-1} \Rightarrow y e = e^{\frac{x}{y}} \Rightarrow y e^{-\frac{x}{y}} = e^{-1}$.Actually,from $\frac{x}{y} = \ln y + 1$,we have $\ln y = \frac{x}{y} - 1$,so $y = e^{\frac{x}{y} - 1} = \frac{e^{x/y}}{e}$.Thus,$y e = e^{x/y}$,which is $y e^{-\frac{x}{y}} = e^{-1}$ or $y e^{1 - x/y} = 1$. Checking the options,$y e^{x/y} = e$ is the standard form derived from $y = e^{x/y - 1} \cdot e = e^{x/y}$.

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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