Show that the line x/a + y/b = 1,

Given curve equation, y = b . e-x/a and line equation x/a + y/b = 1

Now, let the coordinates of the point where the curve intersects the y-axis be (0, y1)

Now differentiating y = b . e-x/a both sides w.r.t. x, we get

$\frac{d y}{d x}=b \cdot e^{-x / a}\left(-\frac{1}{a}\right)=-\frac{b}{a} \cdot e^{-x / a}$

So, the slope of the tangent, $m_{1}=-\frac{b}{a} e^{-x / a}$.

Differentiating $\frac{x}{a}+\frac{y}{b}=1$ both sides w.r.t. $x$, we get

$\frac{1}{a}+\frac{1}{b} \cdot \frac{d y}{d x}=0$

So, the slope of the line, $m_{2}=\frac{-b}{a}$.

If the line touches the curve, then $m_{1}=m_{2}$

$\frac{-b}{a} \cdot e^{-x / a}=\frac{-b}{a} \Rightarrow e^{-x / a}=1$

$\frac{-x}{a} \log e=\log 1$ (Taking log on both sides)

$\frac{-x}{a}=0 \quad \Rightarrow \quad x=0$

Putting $x=0$ in equation $y=b \cdot e^{-x / a}$

$\Rightarrow y=b \cdot e^{0}=b$

$\frac{d y}{d x}=b \cdot e^{-x / a}\left(-\frac{1}{a}\right)=-\frac{b}{a} \cdot e^{-x / a}$

So, the slope of the tangent, $m_{1}=-\frac{b}{a} e^{-x / a}$.

Differentiating $\frac{x}{a}+\frac{y}{b}=1$ both sides w.r.t. $x$, we get

$\frac{1}{a}+\frac{1}{b} \cdot \frac{d y}{d x}=0$

So, the slope of the line, $m_{2}=\frac{-b}{a}$.

If the line touches the curve, then $m_{1}=m_{2}$

$\frac{-b}{a} \cdot e^{-x / a}=\frac{-b}{a} \Rightarrow e^{-x / a}=1$

$\frac{-x}{a} \log e=\log 1$ (Taking log on both sides)

$\frac{-x}{a}=0 \quad \Rightarrow \quad x=0$

Putting $x=0$ in equation $y=b \cdot e^{-x / a}$

$\Rightarrow y=b \cdot e^{0}=b$

Therefore, the equation of the curve intersects at (0, b) which is on the y-axis.