Find the equation of the normal to the curve y = (sin 2x + cot x + 2)2

1.	Find the equation of the normal to the curve y = (sin 2x + cot x + 2)2 at x = π/2.
Answer» Consider y = (sin 2x + cot x + 2)² as the equation of the curve By substituting the value of x = π/2 we get y = 4 On differentiation of both sides w.r.t. x dy/dx = 2(sin 2x + cot x + 2) (2 cos 2x – cosec² x) We get (dy/dx)_x= _π/2 = – 12 Here the equation of normal at point (π/2, 4) y – 4 = 1/12 (x – π/2) On further calculation 12y – 48 = x – π/2 We get 24y – 96 = 2x – π It can be written as 24y – 96 – 2x + π = 0 24y – 2x + π – 96 = 0

Find the equation of the normal to the curve y = (sin 2x + cot x + 2)2 at x = π/2.

Answer»

Consider y = (sin 2x + cot x + 2)² as the equation of the curve

By substituting the value of x = π/2 we get y = 4

On differentiation of both sides w.r.t. x

dy/dx = 2(sin 2x + cot x + 2) (2 cos 2x – cosec² x)

We get

(dy/dx)_x= _π/2 = – 12

Here the equation of normal at point (π/2, 4)

y – 4 = 1/12 (x – π/2)

On further calculation

12y – 48 = x – π/2

We get

24y – 96 = 2x – π

It can be written as

24y – 96 – 2x + π = 0

24y – 2x + π – 96 = 0