Let C be the curve y^3-3 x y+2=0. If H is the set of points on the cu…

We have,

$y^3-3 x y+2=0$            .....(i)

$\Rightarrow 3 y^2 \frac{d y}{d x}-3\left(x \frac{d y}{d x}+y\right)=0$

$\Rightarrow \frac{d y}{d x}=\frac{y}{y^2-x}$

If the tangent is parallel to x-axis, then

$\frac{d y}{d x}=0 \Rightarrow \frac{y}{y^2-x}=0 \Rightarrow y=0$

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

$\frac{d x}{d y}=0 \Rightarrow \frac{y^2-x}{y}=0 \Rightarrow y^2=x$

Putting $x=y^2$ in (i), we get

$y^3-3 y^3+2=0 \Rightarrow y^3=1 \Rightarrow y=1$

∴   $x=y^2 \Rightarrow x=1$

Thus, at (1, 1) the tangent is parallel to y-axis.

∴  $V=\{(1,1)\}$

Hence, $H=\phi$ and $V=\{(1,1)\}$