Find the slopes of the tangent and the normal to the curve `x=a(theta-sintheta)` , `y=a(1+costheta)` at `theta=-pi//2`

Answer

Option (C)

Explanation

The slope of normal ` x=a(\theta-\sin \theta) ; \quad y=a(1-\cos \theta)` at `\theta=\frac{\pi}{2}`



`(\frac{dx}{d \theta})_{\theta=\frac{\pi}{2}}=a(1-\cos \theta)=a`


`(\frac{dy}{d \theta})_{\theta=\frac{\pi}{2}}=a \sin \theta=a`



`\therefore (\frac{dy}{dx})=1`


`\therefore` Slope of normal at the given point is, `m = (-\frac{dx}{dy}) =-1`


Hence, Option ‘C’ is correct.