Prove that all normals to the curve `x=acost+a tsint ,\ \ y=asint-a tcost` are at a distance `a` from the origin.

Answer

Proved

Explanation

Given, `x=a \cos t+a t \sin t \ldots(1)` and `y=a \sin t-a t \cos t \ldots(2)`


Now when differentiating

or, `\frac{d x}{d t}=a(t \cos t)` and `\frac{d y}{d t}=a(t \sin t)`


`\therefore \frac{dy}{dt} = at \tan t`


Slope of tangent `\frac{d y}{d x}=\tan t`


Now normal to the curve is represented by the equation (1) and (2) at the point `[x(t), y(t)]` is
$$
(Y-y(t))=-\left.\frac{d x}{d y}\right|_{[x(t) y(t)]}(X-x(t))
...