The equation of tangent to the curve `y=be^(-x//a)` at the point where it crosses Y-axis is

Options

A. `ax+by=1`
B. `ax-by=1`
C. `x/a-y/b=1`
D. `x/a+y/b=1`

Answer

D

Explanation

Given equation of curve is
`y=be^(-x//a)`…………i
Since the curve crosses Y-axis (i.e. `x=0`)
`:.y=be^(-0)impliesy=b`
On differentiating Eq. (i) w.r.t `x` we get
`(dy)/(dx)=(-b)/a e^(-x//a)`
At point `(0,b),((dy)/(dx))_((0,b))`
`=(-b)/a e^(-0//a)=(-b)/a`
`:.` Required equation of tangen is
`y-b=(-b)/a(x-0)`
`impliesy/b-1=-x/aimpliesx/a+y/b=1`