Question

At which point the line $\frac{x}{a} + \frac{y}{b} = 1$ touches the curve

$y = be^{- x/a}$

• A. (0, 0)
• B. (0, a)
• C. (0, b)
• D. (b, 0)

Answer 1

Answer: (C)

(0, b)

Answer 2

Let the point be $(x_{1},y_{1})$ $\therefore y_{1} = be^{- x_{1}/a}$ ......(i)

Also, curve $y = be^{- x/a} \Rightarrow \frac{dy}{dx} = \frac{- b}{a}e^{- x/a}$

$\left( \frac{dy}{dx} \right)_{(x_{1},y_{1})} = \frac{- b}{a}e^{- x_{1}/a} = \frac{- y_{1}}{a}$ (by (i))

Now, the equation of tangent of given curve at point $(x_{1},y_{1})$ is $y - y_{1} = \frac{- y_{1}}{a}(x - x_{1})$ ⇒ $\frac{x}{a} + \frac{y}{y_{1}} = \frac{x_{1}}{a} + 1$

Comparing with $\frac{x}{a} + \frac{y}{b} = 1$, we get, $y_{1} = b$ and $1 + \frac{x_{1}}{a} = 1$

⇒ $x_{1} = 0$

Hence, the point is (0, b).