### Q. Show that the normal at any point \Theta to the curve

x= acos\theta + a \theta sin\theta , y = a sin\theta-a\theta cos\theta

is at a constant distance from the origin.

Ans:

Note:- This question needs prior knowledge of

The perpendicular distance from a point to the line

i.e.

\therefore AL = \frac{|Ax_1 + By_1 + C|}{\sqrt(A^2 + B^2)}

Now,

x= acos\theta + a \theta sin\theta

\frac{dx}{d\theta} = -a sin\theta + a(sin\theta + \theta cos\theta)

or \frac{dx}{d\theta} = a\theta cos\theta

Also,

y = a sin\theta-a\theta cos\theta

\frac{dy}{d\theta} = a cos\theta - a(cos\theta + (-sin\theta) \theta)

\frac{dy}{d\theta} = a \theta sin\theta

thus,

\frac{dy}{dx} = \frac{a\theta sin\theta}{a \theta cos\theta}

or \frac{dy}{dx} = tan\theta

\therefore slope of normal be = \frac{-1}{tan\theta}

Now,

Equation of normal at (acos\theta + a \theta sin\theta, a sin\theta-a\theta cos\theta)

y - a sin\theta+a\theta cos\theta = \frac{-1}{tan\theta}( x - acos\theta - a \theta sin\theta)

or y tan\theta - a sin\theta tan\theta +a\theta cos\theta tan\theta = -x + acos\theta + a \theta sin\theta)

or y tan\theta - a sin\theta tan\theta +a\theta sin\theta = -x + acos\theta + a \theta sin\theta

or y tan\theta + x = \frac{a sin^2\theta}{cos\theta} + a cos\theta

or y tan\theta + x = \frac{a sin^2\theta +a cos^2\theta}{cos\theta}

or y tan\theta cos\theta + x cos\theta = a

or y sin\theta + x cos\theta - a =0

thus,

Distance of curve from origin is,

= \frac{|sin\theta .0 + cos\theta .0 - a|}{\sqrt(sin^2\theta + cos^2\theta)}

= |-a|

= a

here distance is independent of \theta and equal to a which is a constant.