A curve is represented parametrically by the equations `x=t+e^(at) and y=-t +e^(at, t in R and a gt 0`. If the curve touches the axis of x at the point A, then the coordinates of the point A are

Options

A. (1,0)
B. `((1)/(e,0))`
C. `(e,0)`
D. `(2e,0)`

Answer

D

Explanation

`X=t+e^(at),y=-t+e^(at)`
`(dx)/(dt)=1+ae^(at),(dy)/(dx)=(-1+ae^(at))/(1+ae^(at))`
At the point `A,y=0` and `(dy)/(dx)=0` for some `t=t_(1)`
`becauseae^(at_(1))=1`
(i). Also `0=-t+e^(at_(1))`
`t_(1)=e^(at_(1))`
On putting this value in Eq. (i) we get
`at_(1)=1impliest_(1)=(i)/(a)`
Now from eq `(i), `ae=1impliesa=(1)/(e)`
Hence, `x_(A)=t_(1)+e^(at_(1))=e+e=2e`
`impliesA=(2e,0)`