Question

The equation of a line touching the curve $y={\mathrm{be}}^{-\frac{x}{a}}$ at a point where it crosses the y-axis is .

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

Answer 1

The correct option is A $\frac{x}{a}+\frac{y}{b}=1$

The equation of the given curve is $y={\mathrm{be}}^{-\frac{x}{a}}$ .....[1]
It crosses the y-axis at the point, where x = 0. Putting x = 0 in [1] we get: y = b.
So, the point of contact is (0, b).
Differentiating [1] w.r.t. x, we get
$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{be}}^{-\frac{x}{a}}\frac{d}{\mathrm{dx}}\left(-\frac{x}{a}\right)\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{b}{a}{e}^{-\frac{x}{a}}\Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(0,b)}=-\frac{b}{a}{e}^0=-\frac{b}{a}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{at}\left(0,b\right)\mathrm{is}y-b={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(0,b)}\left(x-a\right)\Rightarrow y-b=-\frac{b}{a}\left(x-0\right)\Rightarrow \mathrm{ay}-\mathrm{ab}=-\mathrm{bx}\Rightarrow \mathrm{bx}+\mathrm{ay}=\mathrm{ab}\Rightarrow \frac{x}{a}+\frac{y}{b}=1.$