The line bx + ay = ab will touch the curve y | vedclass

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(C) The equation of the given line is $bx + ay = ab$,which can be rewritten as $y = -\frac{b}{a}x + b$. The slope of this line is $m = -\frac{b}{a}$.T

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$(0, b)$

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(C) The equation of the given line is $bx + ay = ab$,which can be rewritten as $y = -\frac{b}{a}x + b$. The slope of this line is $m = -\frac{b}{a}$.The equation of the curve is $y = b \cdot e^{-x/a}$.To find the slope of the tangent to the curve at any point $(x, y)$,we differentiate with respect to $x$:$\frac{dy}{dx} = b \cdot e^{-x/a} \cdot \left(-\frac{1}{a}ight) = -\frac{b}{a} e^{-x/a}$.Since the line touches the curve,the slope of the line must be equal to the slope of the tangent at the point of contact:$-\frac{b}{a} = -\frac{b}{a} e^{-x/a}$.Dividing both sides by $-\frac{b}{a}$ (assuming $a, b 
eq 0$),we get:$1 = e^{-x/a}$.Since $e^0 = 1$,we have $-x/a = 0$,which implies $x = 0$.Substituting $x = 0$ into the equation of the curve $y = b \cdot e^{-x/a}$:$y = b \cdot e^0 = b \cdot 1 = b$.Thus,the point of contact is $(0, b)$.

The line $bx + ay = ab$ will touch the curve $y = b \cdot e^{-x/a}$ at the point:

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