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Question

Show that the equation of normal at any point t on the curve x=3costcos3t and y=3sintsin3t is 4(ycos3txsin3t)=3sin4t

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Solution

At a point t, the slope dxdy=dxdt×dtdy
=1×(3sint+3cos2t×sint)×1(3cost3sin2tcost)
=sint×sin2t×1cos3t
=tan3t
Now, the equation of the line be y=mx+c, where m=tan3t
Since the line passes through the given point, we substitute the given values and get 3sintsin3t=3sint×tan2tsin3t+c
c=3sint×cos2t×sec2t
The equation becomes y=tan3tx+3sint×cos2t×sec2t
Multiplying throughout by cos3t, we get ycos3txsin3t=3sint×cos2t×cost
i.e. 4(ycos3txsin3t)=3sin4t

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