show that the equation of normal at any point on the curve x=3 cos a - cos​3 a , y= 3 sin a - sin3a is 4(y cos3a - x sin3a)=3 sin(4a)?



Dear Student,
Please find below the solution to the asked query:

Given curve is:x=3cosa-cos3ay=3sina-sin3adifferentiating w.r.t adxda=-3sina-3cos2a-sina=-3sina1-cos2a=-3sin3adyda=3cosa-3sin2acosa=3cosa1-sin2a=3cos3adydx=dydadxda=3cos3a-3sin3aslope of tangent=-cos3asin3a slope of normal is given by=-1dydx=sin3acos3athe equation of normal to the given point:y-3sina-sin3a=sin3acos3ax-3cosa-cos3aycos3a-3sinacos3a+sin3acos3a=xsin3a-3cosasin3a+sin3acos3aycos3a-xsin3a=3sinacosacos2a-sin2aycos3a-xsin3a=32sin2acos2a=34sin4a4ycos3a-xsin3a=3sin4a

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards
 

  • 40
What are you looking for?