Find the equation of the normal to the curve y=1+sinx/cosx at x= pi/4.

if the given curve is ; i.e.

the slope of the tangent

at x= , ..........(2)

.......(3)

the slope of the normal at x= is

therefore the equation which passes through the point and having slope -1/2 is given as:

which is the required equation of the normal.

hope this helps you.

  • 3

dy/dx=sec2

at x=pi/4  dy/dx=2 and y=2(frm the eqn)

then subs in

y-y1=m(x-x1)

here m=-dx/dy

=> y-2=(-1/2)(x-pi/4)

2y + x=(pi/4) + 4

  • -2
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