Question

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

• A. $(0,a)$
• B. $\left(0.0\right)$
• C. $(0,b)$
• D. $(b,0)$

Answer 1

The correct option is C $(0,b)$

$y=b\times e^{-\frac{x}{a}}$

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

Now the slope of given line is, $m=-\frac{1}{a}\times \frac{b}{1}=-\frac{b}{a}$

Thus for the given line to be tangent to the given curve,

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

$\Rightarrow e^{\frac{x}{a}}=1$

$\Rightarrow x=0\Rightarrow y=b$

Therefore, the point is $(0,b)$

Hence, option 'C' is correct.