Q. The line $\frac{x}{a} + \frac{y}{b} = 1$ touches the curve $y = be^{- x/a}$ at the point
A) $\left( a,\frac{b}{a} \right)$
B) $\left( - a,\frac{b}{a} \right)$
C) $\left( a,\frac{a}{b} \right)$
D) None of these
Explanation
(d)
$\because\ \ y = be^{- x/a}$ …(i)
$\therefore\ \frac{dy}{dx} = - \frac{b}{a}\ e^{- x/a} = - \frac{b}{a}\ \ \ \left( slope\ of\frac{x}{a} + \frac{y}{b} = 1 \right)$
$\Rightarrow \ \ \ e^{- x/a} = 1 = e^{0}$
$\therefore\ \ x = 0$
From Eq. (i), $y = b$
Hence, point of contact is $(0,\ b)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.