Question

The equation of the tangent to curve $y=b{e}^{\frac{-x}{a}}$ at the point where it crosses $y$ – axis is

• A. $ax+by=1$
• B. $ax-by=1$
• C. $\frac{x}{a}–\frac{y}{b}=1$
• D. $\frac{x}{a}+\frac{y}{b}=1$

Answer 1

The correct option is D

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

Explanation for the correct answer:

Step 1: Find the first order derivative of the equation of the curve

Given equation of curve is $y=b{e}^{\frac{-x}{a}}$

Differentiating with respect to $x$ we get

$\Rightarrow$$\frac{dy}{dx}=b\times \frac{-1}{a}\times e^{\frac{-x}{a}}$

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

Step 2: Find the slope of the tangent at required point

Let $P(0,y_1)$ be the point where tangent to curve crosses $y$-axis .

$\Rightarrow y_1=b{e}^0$

$\Rightarrow y_1=b$

So, the point$P$ is$(0,b)$

The first order derivative at a point gives the slope of the tangent at the given point

$\Rightarrow$${\frac{dy}{dx}}_{\left(0,b\right)}=m=\frac{-b}{a}\times e^0$

$\Rightarrow$ $m=\frac{-b}{a}$

Slope of tangent at $P(0,b)$ is $\frac{-b}{a}$

Step3: Write the equation of the tangent using slope point form

Equation of tangent through $P(0,b)$ is

$$\begin{gathered} y-b=\frac{-b}{a}(x-0) \\ \Rightarrow bx+ay=ab \\ \Rightarrow \frac{x}{a}+\frac{y}{b}=1 \end{gathered}$$

Hence, the correct answer is option (D).