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

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(C) Given the curve $y = be^{-x/a}$.To find the point of contact,we first find the derivative of the curve: $\frac{dy}{dx} = b \cdot e^{-x/a} \cdot (-

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$(0, b)$

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(C) Given the curve $y = be^{-x/a}$.To find the point of contact,we first find the derivative of the curve: $\frac{dy}{dx} = b \cdot e^{-x/a} \cdot (-\frac{1}{a}) = -\frac{b}{a} e^{-x/a}$.The given line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $y = -\frac{b}{a}x + b$. The slope of this line is $m = -\frac{b}{a}$.At the point of contact,the slope of the tangent to the curve must equal the slope of the line:$-\frac{b}{a} e^{-x/a} = -\frac{b}{a}$.This implies $e^{-x/a} = 1$,so $-x/a = 0$,which gives $x = 0$.Substituting $x = 0$ into the curve equation $y = be^{-x/a}$,we get $y = be^0 = b$.Thus,the point of contact is $(0, b)$.

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

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