Find the equation of the tangent to the curve | vedclass

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(B) The curve is given by $y = be^{-x/a}$.To find the point where the curve crosses the $y$-axis,we set $x = 0$.Substituting $x = 0$ into the equation

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$\frac{x}{a} + \frac{y}{b} = 1$

Vedclass · Answer

(B) The curve is given by $y = be^{-x/a}$.To find the point where the curve crosses the $y$-axis,we set $x = 0$.Substituting $x = 0$ into the equation,we get $y = be^0 = b$.So,the point of intersection is $(0, b)$.Now,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = b \cdot (-\frac{1}{a}) e^{-x/a} = -\frac{b}{a} e^{-x/a}$.At the point $(0, b)$,the slope $m$ is:$m = (\frac{dy}{dx})_{(0, b)} = -\frac{b}{a} e^0 = -\frac{b}{a}$.The equation of the tangent line passing through $(0, b)$ with slope $m = -\frac{b}{a}$ is given by $y - y_1 = m(x - x_1)$:$y - b = -\frac{b}{a}(x - 0)$.Multiplying by $a$ and rearranging:$a(y - b) = -bx \implies ay - ab = -bx \implies bx + ay = ab$.Dividing both sides by $ab$,we get:$\frac{bx}{ab} + \frac{ay}{ab} = 1 \implies \frac{x}{a} + \frac{y}{b} = 1$.

Find the equation of the tangent to the curve $y = be^{-x/a}$ at the point where it crosses the $y$-axis.

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