Question

The equation of the tangent to the curve $y=4e^{-x/4}$ at the point where the curve crosses Y-axis is equal to

• A. $3x+4y=16$
• B. $4x+y=4$
• C. $x+y=4$
• D. $3x+4y=25$

Answer 1

Answer: (C)

$x+y=4$

As the curve crosses Y-axis i.e. $x=0$

$y=4e^{-0}\Rightarrow y=4$

Given, $y=4e^{-x/4}$

$\Rightarrow \frac{dy}{dx}=4e^{-x/4}(-\frac{1}{4})=-e^{-x/4}$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{0,4}=-e^{-0}=-1$

$\therefore$ Equation of tangent at $(0,4)$ is $y-4=-1(x-0)$

$⟹x+y=4$