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

Answer

5

Explanation

Let the required point be `(x_(1),y_(1))`,
Now, `3y=6-5x^(3)` or `(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))`
Since `(x_(1),y_(1))` lies on the given curve,
`3y_(1)=6x_(1)-5x_(1)^(3)`
Solving equation (1) and (2) we obtain `x_(1)=1` and `y_(1)=1/3`
Hence, the required point is `(1,1/3)` so, `p=1, q=1, r=3`