At which point does the line frac{x}{a} + fra | vedclass

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(C) Let the point of contact be $(x_1, y_1)$.Since the point lies on the curve $y = b e^{-x/a}$,we have $y_1 = b e^{-x_1/a}$ .....$(i)$Now,differentia

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$(0, b)$

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(C) Let the point of contact be $(x_1, y_1)$.Since the point lies on the curve $y = b e^{-x/a}$,we have $y_1 = b e^{-x_1/a}$ .....$(i)$Now,differentiating the curve equation with respect to $x$:$\frac{dy}{dx} = b e^{-x/a} \cdot \left(-\frac{1}{a}\right) = -\frac{b}{a} e^{-x/a}$At the point $(x_1, y_1)$,the slope of the tangent is $m = -\frac{b}{a} e^{-x_1/a} = -\frac{y_1}{a}$ (using equation $(i)$).The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{a}(x - x_1)$.Rearranging this,we get $\frac{y}{y_1} - 1 = -\frac{x}{a} + \frac{x_1}{a}$,which simplifies to $\frac{x}{a} + \frac{y}{y_1} = 1 + \frac{x_1}{a}$.Comparing this with the given line equation $\frac{x}{a} + \frac{y}{b} = 1$,we equate the coefficients:$y_1 = b$ and $1 + \frac{x_1}{a} = 1$.From $1 + \frac{x_1}{a} = 1$,we get $x_1 = 0$.Thus,the point of contact is $(0, b)$.

At which point does the line $\frac{x}{a} + \frac{y}{b} = 1$ touch the curve $y = b e^{-x/a}$?

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