Show that the normal at any point θ to the curve is at a constant distance from the origin.
The given equation of the curve is,
x = a cos θ + a θ sin θ y = a sin θ − a θ cos θ
Differentiate both the equations
d x d θ = a [ − sin θ + θ cos θ + sin θ ] = a θ cos θ d y d θ = a [ cos θ + θ sin θ − cos θ ] = a θ sin θ
Write the expression for the slope of the tangent at the point ( x , y ) on the curve.
d y d x = d y d θ d x d θ = a θ sin θ a θ cos θ = tan θ
Since the slope of the normal is negative of reciprocal of the slope of tangent, therefore write the expression for the slope of the normal at any point θ on the curve.
− 1 tan θ = − cot θ = − cos θ sin θ
The formula for the equation of normal is,
y − y 1 = m ( x − x 1 )
Substitute x 1 = a cos θ + a θ sin θ and y 1 = a sin θ − a θ cos θ in the above equation.
y − a ( sin θ − θ cos θ ) = − cos θ sin θ ( x − a ( cos θ + θ sin θ ) y sin θ − a sin 2 θ + a θ cos θ sin θ = − x cos θ + a cos 2 θ + a θ sin θ cos θ x cos θ + y sin θ = a ( sin 2 θ + cos 2 θ ) x cos θ + y sin θ = a
The formula for the perpendicular distance of the normal from the origin is,
d = | 0 + 0 − a | cos 2 θ + sin 2 θ = a
Thus, the distance of normal from origin ( 0 , 0 ) is constant.
The given equation of the curve is,
Differentiate both the equations
Write the expression for the slope of the tangent at the point
Since the slope of the normal is negative of reciprocal of the slope of tangent, therefore write the expression for the slope of the normal at any point
The formula for the equation of normal is,
Substitute
The formula for the perpendicular distance of the normal from the origin is,
Thus, the distance of normal from origin
