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प्रश्न
Find the equation of the tangent and the normal to the following curve at the indicated points:
x = 3cosθ − cos3θ, y = 3sinθ − sin3θ?
उत्तर
`x=3costheta-cos^3theta,` `y=3sintheta-sin^3theta`
`rArr(dx)/(d theta)=-3sintheta +3cos^2thetasintheta `
And
`rArr(dy)/(d theta)=3costheta-3sin^2thetacostheta`
`rArr(dy)/(dx)=((dy)/(d theta))/((dx)/(d theta))=(3costheta-3sin^2thetacos theta)/(-3sintheta+3cos^2thetasintheta)=(costheta(1-sin^2theta))/(-sintheta(1-cos^2theta))=cos^3theta/-sin^3theta=-tan^3theta`
So equation of the tangent at θ is
`y-3sintheta+sin^3theta=-tan^3theta(x-3costheta+cos^3theta)`
`rArr4(ycos^3theta-xsin^3theta)=3sin4theta`
So equation of normal at θ is
`y-3sintheta+sin^3theta=1/tan^3theta(x-3costheta+cos^3theta)`
`rArrycos^3theta-xcos^3theta=3sin^4theta-sin^6theta-3cos^4theta+cos^6theta`
`rArr ysin^3theta-xcos^3theta=3sin^4theta-sin^6theta-3cos^4theta+cos^6theta`
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