Practicing Success
Let C be the curve $y^3-3 x y+2=0$. If H is the set of points on the curve C where the tangent is parallel to x-axis and V is the set of points on C where the tangent is parallel to y-axis, then |
$H=\{(x, y): y=0, x \in R\}, V=\{(1,1)\}$ $H=\{(x, y): x=0, y \in R\}, V=\{(1,1)\}$ $H=\phi, ~~V=\{(1,1)\}$ $H=\{(1,1)\},~~V=\{(x, y): y=0, x \in R\}$ |
$H=\phi, ~~V=\{(1,1)\}$ |
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 $V=\{(1,1)\}$ |