Question

Consider the curve $y=b{e}^{-x/a}$ where $a$ and $b$ are non-zero real numbers. Then

• A. $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve at $(0,0)$.
• B. $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve where the curve crosses the axis of $y$.
• C. $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve at $(a,0)$
• D. $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve at $(2a,0).$

Answer 1

The correct option is B $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve where the curve crosses the axis of $y$.

$y=b{e}^{-x/a}$
$\Rightarrow \frac{dy}{dx}=\frac{-b}{a}{e}^{-x/a}$

Slope of the tangent $\frac{x}{a}+\frac{y}{b}=1$ is $\frac{-b}{a}$
Also, $\frac{dy}{dx}=\frac{-b}{a}$ when $x=0$
So, $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve at the point where $x=0$
$x=0\Rightarrow y=b$
So, $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve at the point $(0,b)$