Consider the curve y = b e^{-frac{x}{a}},wher | vedclass

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(B) Given the curve $y = b e^{-\frac{x}{a}}$.Find the derivative: $\frac{dy}{dx} = b e^{-\frac{x}{a}} \cdot (-\frac{1}{a}) = -\frac{y}{a}$.At any poin

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$\frac{x}{a} + \frac{y}{b} = 1$ is tangent to the curve where the curve crosses the $y$-axis

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(B) Given the curve $y = b e^{-\frac{x}{a}}$.Find the derivative: $\frac{dy}{dx} = b e^{-\frac{x}{a}} \cdot (-\frac{1}{a}) = -\frac{y}{a}$.At any point $(x_0, y_0)$ on the curve,the slope of the tangent is $m = -\frac{y_0}{a}$.The equation of the tangent line is $y - y_0 = -\frac{y_0}{a}(x - x_0)$.Dividing by $y_0$: $\frac{y}{y_0} - 1 = -\frac{x}{a} + \frac{x_0}{a}$.Rearranging: $\frac{x}{a} + \frac{y}{y_0} = 1 + \frac{x_0}{a}$.For this to match $\frac{x}{a} + \frac{y}{b} = 1$,we must have $y_0 = b$ and $1 + \frac{x_0}{a} = 1$,which implies $x_0 = 0$.At $x_0 = 0$,$y_0 = b e^0 = b$. Thus,the point of tangency is $(0, b)$.Since the curve crosses the $y$-axis at $x = 0$,the tangent at the $y$-intercept is $\frac{x}{a} + \frac{y}{b} = 1$.

Consider the curve $y = b e^{-\frac{x}{a}}$,where $a$ and $b$ are non-zero real numbers. Then:

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