Please help - Equation of the Normal Q. (1 Viewer)

[Utility]

Find the equation of the normal to the curve y = tan(x) at the point (pi /3, √3). Thanks.

 

[Aesytic]

dy/dx = sec^2(x)
dy/dx (when x=pi/3) = sec^2(pi/3)
= 4
.'. gradient of tangent at (pi/3, root3) is 4
.'. gradient of normal at this point is -1/4
using point gradient formula,
y - root3 = -1/4(x-pi/3)
4y - 4root3 = -x + pi/3
x + 4y - (4root3 + pi/3) = 0

 

[Sy123]

 

[Utility]

Nice, thanks guys.