Let C be the curve `y^(3) - 3xy + 2 =0`. If H is the set of points on the curve C, where the tangent is horizontal and V is the set of points on the curve C, where the tangent is vertical, then H = … and V = … .

Answer

`H= phi, V = {1, 1}`

Explanation

Given, `y^(3) - 3xy + 2 =0`
On differentiating w.r.t. x, we get
` 3y ^(2) (dy)/(dx) - 3x (dy)/(dx) - 3y =0`
`rArr (dy)/(dx) ( 3y ^(2) - 3x) = 3y rArr (dy)/(dx) = ( 3y)/( 3y^(2) - 3x)`
For the points where tangent is horizontal, the slope of tangent is zero.
i.e. `" " (dy)/(dx) = 0 rArr ( 3y )/( 3y ^(2) - 3x) = 0`
`rArr y =0` but `y =0` does not satisfy the given equation of the curve, therefore y cannot lie on the curve.
So, `" "H = phi" " ` [null sec]
For the point where tangent is vertical, `(dy)/(dx) = oo`
`rArr " " (y)/( y^(2) - x) =oo`
`rArr " " y^(2) - x =0`
`rArr " " y ^(2) = x `
On putting this value in the given equation of the curve,
we get
`" " y ^(3) - 3 * y^(2) *y + 2 =0`
`rArr " " - 2y ^(3) + 2 =0`
`rArr y ^(3) - 1 =0 rArr y ^(3) = 1 `
`rArr " " y =1, x = 1 `
Then, `" " V = {1, 1}`