Find the equation of a normal to the curve `y = x log_e x` which is parallel to the line `2x - 2y + 3 = 0.`

Options

A. (0,0)
B. (e,e)
C. `(e^(2),2e^(2))`
D. `(e^(-2),-2e^(-2))`

Answer

D

Explanation

Given `y=logx`
On differentiating w.r.t.x we get
`(dy)/(dx)=1+logx`
`:.` Slope of normal `=-(1)/(dy//dx)`
`=(-1)/(1+logx)`
The slope of the given line
`2x-2y=3` is 1.
Also, normal parallel to given line.
`:.-(1)/(1+logx)=1`
`implies1=-1-logx`
`impliesx=e^(-2)`
and `y=-2e^(-2)`
Hence, coordinates of point are
`(e^(-2),-2e^(-2))`