The point on the curve `3y=6x-5x^(3)` the normal at which passes through the orgin is

Options

A. `(1,1//3)`
B. `(1//3,1)`
C. `(2,-28//3)`
D. `(-1,-1//3)`

Answer

A

Explanation

Let the required point be `(x_(1),y_(1))`
Now,
` 3y=6x-5x^(3) `
` rArr 3 (dy)/(dx)=6-15x^(2) rArr (dy)/(dx)= 2-5x^(2) rArr ((dy)/(dx))_((x_(1)","y_(1))) =2-5x_(1)^(2) `
The equation of the normal at `(x_(1),y_(1))` is
`y-y_(1)=(-1)/(2-5x_(1)^(2))(x-x_(1))`
If it passes through the origin, then
`0-y_(1)=(1)/(2-5x_(1)^(2))(0-x_(1)) rArr y_(1)=(-x_(1))/(2-5x_(1)^(2)) " "...(i)`
Since `(x_(1),y_(1))` lies on the given curve. Therefore,
` 3y_(1)=6x_(1)-5x_(1)^(3) " "...(ii) `
Solving (i) and (ii), we obtain `x_(1)=1 " and " y_(1)=1//3`
Hence, the required point is `(1,1//3).`