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

Options

A. `(1//2,1//4)`
B. `(1//4,1//2)`
C. `(4,2)`
D. `(1,1)`

Answer

B

Explanation

Let `(x_(1),y_(1) )` be the required point on the curve `y^(2)=x` Then,
`((dy)/(dx))_((x_(1)","y_(1)))=1`
`rArr (1)/(2y_(1))=1 " "[because y^(2)=x rArr2y(dy)/(dx)=1 rArr(dy)/(dx)=(1)/(2y)]`
`rArr y_(1)=(1)/(2)`
Since `(x_(1),y_(1))` lies on `y^(2)=x.`
`therefore x_(1)=y_(1)^(2)rArr x_(1)=(1)/(4)`
Hence, `(1//4,1//2)` is the required point.