Given curve is y = x ln (e x)
\(\therefore\) slope of tangent to curve is \(\frac{dy}{dx}=\frac{x}{ex}\times e+ln(ex)\)
= 1 + ln(ex) = lne + ln ex
= ln(e2x)
Given line is x + 2y + 3 = 0
slope of perpendicular line to line x + 2y + 3 = 0
= \(\frac{-1}{slope\,of\,line\,x + 2y+3=0}\)
= \(\cfrac{-1}{\frac{-1}2}\) = 2
Since, given that tangent to curve is perpendicular to the line x + 2y + 3 = 0
So, their slopes are equal.
\(\therefore\) ln(e2x) = 2
⇒ e2x = e2
⇒ x = 1
\(\therefore\) y = x ln(ex) = 1 ln(e) = 1
Hence, required point is (1, 1) which the tangent to given curve is perpendicular to the line
x + 2y + 3 = 0