Step 1: Understanding the Question:
Find the tangent equation at $x = 0$ (where the curve crosses the Y-axis). Step 2: Key Formula or Approach:
1. Find point of contact: put $x = 0$ in curve equation.
2. Find slope: evaluate $dy/dx$ at $x = 0$.
3. Line equation: $y - y_1 = m(x - x_1)$. Step 3: Detailed Explanation:
Point of contact: at $x = 0$, $y = b e^0 = b$. Point is $(0, b)$.
Differentiating: $\frac{dy}{dx} = b e^{-x/a} \times \left( -\frac{1}{a} \right) = -\frac{b}{a} e^{-x/a}$.
Slope at $x = 0$: $m = \left. \frac{dy}{dx} \right|_{x=0} = -\frac{b}{a} e^0 = -\frac{b}{a}$.
Equation of tangent:
$y - b = -\frac{b}{a}(x - 0)$
$y - b = -\frac{b}{a}x$
$\frac{b}{a}x + y = b$
Divide the entire equation by $b$:
$\frac{x}{a} + \frac{y}{b} = 1$. Step 4: Final Answer:
The tangent equation is $\frac{x}{a} + \frac{y}{b} = 1$.