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Question
Show that the equation of normal at any point t on the curve x = 3 cos t – cos3t and y = 3 sin t – sin3t is 4 (y cos3t – sin3t) = 3 sin 4t
Solution
Given:
x=3 cost−cos3t
y=3 sint−sin3t
Slope of the tangent, `dy/dx=(dy/dt)/(dx/dt)=(3cost-3sin^2tcost)/(-3sint+3cos^2tsint)`
`=(3cost[cos^2t])/(-3sint[sin^2t])`
`dy/dx=(-cos^3t)/sin^3t`
∴Slope of the normal
`=sin^3t/cos^3 t`
The equation of the normal is given by
`(y-(3sint-sin^3t))/(x-(3cost-cos^3t))=sin^3t/cos^3t`
`=>ycos^3t-3sint cos^3t +sin^3tcos^3t=xsin^3t-3costsin^3t+sin^3tcos^3t`
`=>ycos^3t-xsin^3t=3(sintcos^3t-costsin^3t)`
`=>ycos^3t-xsin^3t=3sintcost(cos^2t-sin^2t)`
`=>ycos^3t-xsin^3t=3/2sin2tcos2t=3/4sin4t`
`=>4(ycos^3t-xsin^3t)=3sin4t`
Hence proved.
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