The slope of the normal to the curve `x=a(theta-sintheta), y=a(1-costheta)` at `theta=pi/2`

The equation of the normal to the curve x=a(theta-sintheta),y=a(1-costheta) at (theta=pi//2) is

The slope of the nomal to the curve x = a(theta - sin theta), y = a(1-cos theta) at theta = pi/2 is

Find the slopes of the tangent and the normal to the curve x=a(theta-sintheta) , y=a(1+costheta) at theta=-pi//2

Find the slopes of the tangent and the normal to the curve x=a(theta-sintheta),\ \ y=a(1-costheta) at theta=pi//2

Find the slopes of the tangent and the normal to the curve x=a(theta-sintheta) , y=a(1+costheta) at theta=-pi//2

Find the slopes of the tangent and the normal to the curve x=a(theta-sintheta),\ \ y=a(1-costheta) at theta=pi//2

Find the slope of the tangent to the curve: x = a(theta-sintheta), y = a(1-costheta) at theta = pi/2

Find the length of normal to the curve x=a(theta+sintheta),y=a(1-costheta) at theta=pi/2dot

Find the length of normal to the curve x=a(theta+sintheta),y=a(1-costheta) at theta=pi/2dot

Find the length of normal to the curve x=a(theta+sintheta),y=a(1-costheta) at theta=pi/2dot