Find the equation of normal to curve x=cos b , y=sin b at the point where b=pi/4 . Find the coordinates of the point on the curve where the normal meets the curve again .
The Attempt at a Solution
Gradient of normal :
gradient of normal = tan b , when b=pi/4 , gradient of normal =1
The normal cuts the curve at point (1/sqrt(2) , 1/sqrt(2))
equation of normal is y-1/sqrt(2)=1(x-1/sqrt(2))
y=x --- 1
The cartesian equation of the curve is cos^(-1) x=sin^(-1) y
sin y=cos x --- 2
combine 1 and 2 , sin x=cos x
x=0 , pi , 2pi
Since the domain is not mentioned , i take it as 0<=x<=2pi
Therefore , the normal would cut the curve at points (0,pi/2) , (pi,3pi/2), (2pi, 5pi/2)
Am i correct ?