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
$\frac{dy}{dx}=b{e}^{\frac{-x}{a}}\left(-\frac{1}{a}\right)$
$=-\frac{b}{a}{e}^{\frac{-x}{a}}.....$(ii)
At $P(0,b),\frac{dy}{dx}=-\frac{b}{a}$.
Now, equation of tangent at $P(0,b)$ is
$y-b=-\frac{b}{a}(x-0)\Rightarrow ay-ab=-bx\Rightarrow bx+ay=ab$
$\Rightarrow \frac{x}{a}+\frac{y}{b}=1$
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.