Question

The coordinates of a point on the curve y = x ln (ex) at which the tangent is perpendicular to the line x + 2y + 3 = 0, are

Answer 1

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