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

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