Let C be the curve `y-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 = ________

Options

A. `H={(x,y):y=0, x inR}, V={(1,1)} `
B. `H={(x,y):x=0, y in R}, V={(1,1)} `
C. `H=phi , V={(1,1)} `
D. `H={(1,1)} , V={(x,y):y=0, x in R} `

Answer

C

Explanation

We have,
` y^(3)-3xy+2=0 " "...(i)`
` rArr 3y^(2)(dy)/(dx)-3(x(dy)/(dx)+y)=0 `
` rArr (dy)/(dx)=(y)/(y^(2)-x) `
If the tangent is parallel to x-axis, then
` (dy)/(dx)=0 rArr (y)/(y^(2)-x)=0 rArr y=0 `
But, `y=0 ` does not satisfy equation (i). So, there is no point on the curve where tangent is parallel to x-axis. Therefore, `H= phi.` For the tangent to be parallel to y-axis, we must have
`(dx)/(dy)=0 rArr (y^(2)-x)/(y)=0 rArr y^(2)=x `
Putting `x=y^(2)` in (i), we get
`y^(3)-3y^(3)+2=0 rArr y^(3)=1 rArr y=1 `
` therefore x=y^(2) rArr x=1 `
Thus, at (1,1) the tangent is parallel to y-axis.
` therefore V={(1,1)} `
Hence, `H= phi " and " V={(1,1)} `