Question

At which point the line $\frac{x}{a}+\frac{y}{b}=1,$ touches the curve $y=b{e}^{-\frac{x}{a}}$

• A. $\left(0,0\right)$
• B. $\left(0,a\right)$
• C. $\left(0,b\right)$
• D. $\left(b,0\right)$

Answer 1

The correct option is C

$\left(0,b\right)$

Explanation for correct answer:

Step-1: Finding slope of the curve.

Let the line touches the curve at point $P\left(x_1,y_1\right)$

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

Differentiate with respect to $x$

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

$\Rightarrow$$\frac{dy}{dx}=-\frac{b}{a}e\left({}^{-\frac{x}{a}}\right)$

Step-2: Finding slope of the line.

Given, equation of the line is $\frac{x}{a}+\frac{y}{b}=1,$

Slope of the line is $\left(-\frac{\left({\displaystyle \frac{1}{a}}\right)}{\left({\displaystyle \frac{1}{b}}\right)}\right)$

Step-3: Equate slope of line and curve.

$\Rightarrow$$\left(-\frac{\left({\displaystyle \frac{1}{a}}\right)}{\left({\displaystyle \frac{1}{b}}\right)}\right)=-\frac{b}{a}e\left({}^{-\frac{x}{a}}\right)$

$\Rightarrow$ $\left(-\frac{b}{a}\right)=-\frac{b}{a}e\left({}^{-\frac{x}{a}}\right)$

$\Rightarrow$ $1=e\left({}^{-\frac{x}{a}}\right)$

$\Rightarrow$ $e^0=e\left({}^{-\frac{x}{a}}\right)$

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

$\Rightarrow$ $x=0$

Put value of $x$ in equation of curve.

$\Rightarrow$ $y=b{e}^{-\frac{0}{a}}$

$\Rightarrow$ $y=b$

Therefore, required point is $\left(0,b\right)$

Hence, option $\left(C\right)$ is correct.