Find the equations of the tangent and the normal at the point `' t '` on the curve `x=a\ sin^3t` , `y=b\ cos^3t` .

Options

A. `4CT^(2)=ON^(2)=a^(2)`
B. the length of the tangent `= |(y)/(cos t)|`
C. the length of the normal ` = |(y)/(sint)|`
D. all the above

Answer

C

Explanation

At any point 't ' on the given curve, we have
`(dy)/(dx) = (dy)/(dt)/(dx)/(dt)=(3a cos^(2)t(-sint))/(3a sin^(2)t(-cost))= -cot t `
The equation of the tangent and normal at ' t ' are
`x cos t + y sint -(a)/(2) sin 2t =0 " " `..(i)
`and, xsint " "ycost + a cos2t =0 " " ` ...(ii)
`therefore OZ=(|(a//2)sin2t|)/(sqrt(cos^(2)t+sin^(2)t))=(a)/(2) sin 2t rArr 2 OT = a sin2t`
`and,ON=(|a cos 2t|)/(sqrt(sin^(2)t+cos^(2)t))=acos2t`
Hence,
`4OT^(2)+ON^(2)=a^(2) sin^(2) 2t+a^(2)cos^(2)2t=a^(2)`
Length of the tangent at 't' `= (|y sqrt(1+((dy)/(dx))^(2))|)/(dy//dx)`
`rArr " Length of the tangent at 't'" =|(y "cosec"t)/(-cot t)|=((y)/(cost))`
Length of the normal at 't' `= |y sqrt(1+((dy)/(dx))^(2))|`
`rArr" Length of the normal at 't'" = |y sqrt(1+cot^(2)t)|=|(y)/(sin t)|`
Hence, options (a), (b) and (c) are correct.