Find the slope of the normal to the curve `x=1-asintheta` , `y=b\ cos^2theta` at `theta=pi/2` .

Explanation

`x = 1 - a sin theta`
`rArr (dx)/(d theta) =- a cos theta`
` y = cos^(2) theta`
` rArr (dy)/(d theta) = b* 2 cos theta d/(d theta) 9-cos theta)`
` =- 2b cos theta sin theta`
`:. (dy)/(dx) = ((dy)//(d theta))/((dx)//(d theta)) = (-2 cos theta sin theta)/(-a cos theta) = (2b)/a sin theta`
at theta = pi/2`
Slope of tangent
` m = (2b)/a sin pi/2 = (2b)/a`
and slope of normal = `- 1/m =-a/(2b)`