The normal to the curve `x=a(cos theta + theta sin theta), y=a(sin theta - theta cos theta)` at any `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 that, `x=a(costheta+thetasintheta)`
`(dx)/(d theta)=a(-sintheta+sintheta+thetacostheta)`
`implies" "(dx)/(d theta)=athetacostheta`
and `" "y=a(sintheta-thetacostheta)`
`(dy)/(d theta)=a(costheta-costheta+thetasintheta)`
`implies" "(dy)/(d theta)=athetasintheta`
`:." "(dy)/(dx)=((dy)/(d theta))/((dx)/(d theta))`
`=tan theta`
Slope of normal `=-(dx)/(dy)`
`=-cot theta`
So, equation of normal is
`y-a sin theta+a theta cos theta`
`=-(cos theta)/(sintheta)(x-a cos theta-a thetasintheta)`
`impliessinthetay-asin^(2)theta+athetacosthetasintheta`
`=-xcostheta+acos^(2)theta+athetasinthetacostheta`
`implies" "xcostheta+ysintheta=a`
It is always at a constant distance a from origin.