Explanation

(d)
Given equation of curve is$y = be^{- \frac{x}{a}}$
Since, the curve crosses $y$-axis $ie,\ x = 0$
$\Rightarrow \ \ \ y = be^{- 0}\ \ \ \Rightarrow \ \ y = b$
On differentiating Eq. (i) w.r.t. $x$, we get
$\frac{dy}{dx} = \frac{- b}{a}\ e^{- x/a}$
At point $(0,\ b),\ \left( \frac{dy}{dx} \right)_{(0,\ b)} = \frac{- b}{a}e^{- 0/a} = \frac{- b}{a}$
$\therefore$ Required equation of tangent is
$y - b = \frac{- b}{a}(x - 0)$
$\Rightarrow \ \ \frac{y}{b} - 1 = - \frac{x}{a}$
$\Rightarrow \ \ \frac{x}{a} + \frac{y}{b} = 1$