# Show that the line $\frac{x}{a}+\frac{y}{b}=1$ touches the curve $\mathrm{y}=\mathrm{b.e}^{-\frac{x}{a}}$ at the point where the curve intersects the axis of y.

Given: equation of line $\frac{x}{a}+\frac{y}{b}=1$ the curve $\mathrm{y}=\mathrm{b.e}^{-\frac{x}{a}}$ intersects the y-axis

To show: the line touches the curve at the point where the curve intersects the axis of y

Explanation: given the curve $\mathrm{y}=\mathrm{b.e}^{-\frac{x}{a}}$ intersects the y-axis, i.e., at x = 0

Now differentiate the given curve equation with respect to x, i.e.,

$\frac{d y}{d x}=\frac{d\left(b \cdot e^{-\frac{x}{a}}\right)}{d x} \\Removing all the constant term, \\\Rightarrow \frac{d y}{d x}=b \frac{d\left(e^{-\frac{x}{a}}\right)}{d x} \\After differentiate in the equation we get \\\Rightarrow \frac{d y}{d x}=b \cdot e^{-\frac{x}{a}} \frac{d\left(-\frac{x}{a}\right)}{d x} \\\Rightarrow \frac{d y}{d x}=b \cdot e^{-\frac{x}{a}} \cdot\left(-\frac{1}{a}\right) \\Now put the value of x=0, \\\Rightarrow\left(\frac{d y}{d x}\right)_{x=0}=b \cdot e^{-\frac{0}{a}} \cdot\left(-\frac{1}{a}\right) \\\Rightarrow\left(\frac{d y}{d x}\right)_{x=0}= b. 1 \cdot\left(-\frac{1}{a}\right) \\\Rightarrow\left(\frac{d y}{d x}\right)_{x=0}=\left(-\frac{b}{a}\right)=m_{1} \ldots \ldots(i)$

Then considering the line equation,

$\frac{x}{a}+\frac{y}{b}=1 \\Now differentiate it with respect to X \\\frac{\mathrm{d}\left(\frac{\mathrm{X}}{\mathrm{a}}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{b}}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(1)}{\mathrm{dx}} \\Removing all the constant terms \\\Rightarrow \frac{1}{a}+\frac{1}{b} \frac{d y}{d x}=0 \\\Rightarrow \frac{1}{\mathrm{~b}} \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{\mathrm{a}} \\\Rightarrow \frac{d y}{d x}=-\frac{b}{a}=m_{2} \ldots .$

Line touches the curve only if their slopes are equal

From equation (i) and (ii), we see that

$m_1 = m_2$

Hence, the line touches the curve at the point where the curve intersects the axis of y.