Prove that the segment of the normal to the curve `x= 2a sin t + a sin t cos^2 t ; y = - a cos^3 t` contained between the co-ordinate axes is equal to `2a.`

Options

A. normal is inclined at an angle `(pi)/(2)+t` with x-axis.
B. normal is inclined at an angle t with x-axis.
C. portion of normal contained between the co-ordinate axes is equal to 2a.
D. portion of normal containned between the co-ordinate axes is equal to 4a.

Answer

A, C

Explanation

`(dy)/(dx)=(dy//dt)/(dx//dt)=(3acos^(2)tsint)/(2a cost+a cos^(3)t-2a sin^(2)t cost)`
`=(3acos^(2)t sin t)/(2acost(1-sin^(2)t)+a cos^(3)t)`
`=(3acos^(2)t sin t)/(2a cos^(3)t+a cos^(3)t)`
`=(3 a cos^(2)t sin t)/(3acos^(3)t)`
`(dy)/(dx)=tant`, equation of normal is
`y+a cos^(3)t=-(cott)/(sint)(x-2a sin t - a sin t cos^(2)t)`
`therefore" "y sin t+a cos^(3) t sin t=-x cos t+2a sin t cos t+a sin t cos^(3)t`
`x cos t+y sin t=2a sin t cos t`
It meets axis at `A(2a sin t,0) and B(0, 2a cos t)`
Then AB = 2a.