Given equation of curve is \\\\(y=be^{-x/a}\\\\)Let \\\\(P(0,y_1)\\\\) be the point where tangent to curve crosses y-axis .\\\\(\\Rightarrow y_1=b\\\\)So, the point P is (0,b)\\\\(\\dfrac{dy}{dx}=-\\dfrac{y}{a}\\\\)Slope of tangent at P(0,b) is \\\\(-\\dfrac{b}{a}\\\\)Equation of tangent through P(0,b) is \\\\(y-b=-\\dfrac{b}{a}(x-0)\\\\)\\\\(\\Rightarrow bx+ay=ab\\\\)\\\\(\\dfrac {x}{a}+\\dfrac{y}{b} =1\\\\)