Question

Equation of tangent at the curve $y=b e^{-x / a}$ crosses the $Y$-axis is

• A. $\frac{x}{a}+\frac{y}{b}=1$
• B. $\frac{x}{a}-\frac{y}{b}-1$
• C. $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1$
• D. $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Answer 1

Answer: (A)

Given curve is $y=b e^{\frac{-x}{a}} ....$(i)
At the point where curve crosses the $Y$-axis, $x=0$
$\therefore$ From Eq. (i), $y=b e^0=b$
Thus, curve crosses the $Y$-axis at point $P(0, b)$.
On differentiating Eq. (i), we get
$\frac{d y}{d x} =b e^{\frac{-x}{a}}\left(-\frac{1}{a}\right) $
$ =-\frac{b}{a} e^{\frac{-x}{a}} .....$(ii)
At $P(0, b), \frac{d y}{d x}=-\frac{b}{a}$.
Now, equation of tangent at $P(0, b)$ is
$ y-b=-\frac{b}{a}(x-0) \Rightarrow a y-a b=-b x \Rightarrow b x+a y=a b $
$ \Rightarrow \frac{x}{a}+\frac{y}{b}=1$
Hence, line $\frac{x}{a}+\frac{y}{b}=1$ touches the curve $y=b e^{\frac{-x}{a}}$ at the point where it crosses the $Y$-axis.