The normal to the curve `x=a ( cos theta-thetasintheta), y = a( sintheta-theta costheta)` at any point , `theta` , is such that

Options

A. it is at a constant distance from the origin
B. it passes through `((a pi)/2,-a)`
C. It makes angle `(pi)/2-theta` with the X-axis
D. It passes through the origin.

Answer

A

Explanation

Given `x=a(cos theta+theta sin theta)`
and `y=a(sin theta-thetacos theta)`
`implies (dx)/(d theta)=a(-sin theta+ sin theta+ theta cos theta)`
`implies (dx)/(d theta)=a theta cos theta`
and `(dy)/(d theta)=a( cos theta - cos theta+ theta sin theta)`
`implies (dy)/(d theta)=a theta sin theta :. (dy)/(dx)=tan theta`
Then, slope of normal `=(-1)/(tan theta)=-(cos theta)/(sin theta)`
So equation of normal is
`y-a sin theta+ a theta cos theta=-(cos theta)/(sin theta)(x-a cos theta -a theta sin theta)`
`impliesy sin theta -a sin^(2) theta +a theta cos theta sin theta`
`=-x cos theta + a cos^(2) theta+ a theta sin theta cos theta`
`implies x cos theta +y sin theta =a`
It is always at a constant distance 'a' from origin.