The equation of the tangent to the curve `y=4e^(-x/4)` at the point where the curve crosses Y-axis is equal to

Options

A. `3x+4y=16`
B. `4x+y=4`
C. `x+y=4`
D. `4x-3y=-12`

Answer

C

Explanation

Given `y=4e^(-x//4)`
As the curve crosses Y-axis, so `x=0`
`:.y=4e^(-0)impliesy=4`
Now `(dy)/(dx)=4e^(-x/4)(-1/4)=-e^(-x/4)`
At `(0,4),((dy)/(dx))_((0,4))=-e^(-0)=-1`
`:.` Equation of tangent at (0,4) is
`y-4=-1(x-0)impliesx+y=4`