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 = … .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} `

Let C be the curve `y^(3) - 3xy + 2 =0`. If H is the set of points on

1.	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 = … .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» Correct Answer - C 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)} `

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