Question

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 where the tangent is vertical, then $H$ and $V$ are respectively given by :

• A. $\left\{\left(0,0\right)\right\},\left\{\left(0,1\right)\right\}$
• B. $\phi ,\left\{\left(1,1\right)\right\}$
• C. $\left\{\left(1,1\right)\right\},\left\{\left(0,0\right)\right\}$
• D. None of these

Answer 1

The correct option is B

$\phi ,\left\{\left(1,1\right)\right\}$

Explanation for the correct option:

Given, $y^3-3xy+2=0$

Step 1 : Differentiate the given equation with respect to $x$

$$\begin{gathered} \Rightarrow 3y^2\frac{dy}{dx}-3x\frac{dy}{dx}+3y=0 \\ \Rightarrow \frac{dy}{dx}\left(3y^2-3x\right)=3y \\ \Rightarrow \frac{dy}{dx}=\frac{3y}{3y^2-3x} \\ \Rightarrow \frac{dy}{dx}=\frac{y}{y^2-x} \end{gathered}$$

Step 2: Point Where tangent is horizontal

For a point where the tangent is horizontal, the slope of the tangent is zero, i.e : $\frac{dy}{dx}=0$

Then we can say that $\frac{y}{y^2-x}=0\left[\because \frac{dy}{dx}=\frac{y}{y^2-x}\right]$

$\therefore y=0$ but if we put $y=0$, it does not satisfy the equation of a curve, therefore $y$ cannot lie on the curve

hence, $H$ is the empty set $H=\phi$ [null set]

Step 3: Point where the tangent is vertical

For a point where the tangent is vertical, the slope of the tangent is zero, i.e : $\frac{dy}{dx}=\infty$

$$\begin{gathered} \therefore \frac{y}{y^2-x}=\infty \left[\because \frac{dy}{dx}=\frac{y}{y^2-x}\right] \\ {y}^2-x=0 \\ {y}^2=x \end{gathered}$$

Step 4: Substitute the obtained value of $y$ in the given equation of a curve

$$\begin{gathered} \Rightarrow y^3-3xy+2=0 \\ \Rightarrow y^3-3.y^2.y+2=0 \\ \Rightarrow -2y^3+2=0 \\ \Rightarrow y^3-1=0 \\ \Rightarrow y^3=1 \end{gathered}$$

That implies :

$$\begin{gathered} y=1 \\ x=1 \end{gathered}$$

Then, $V=\left\{1,1\right\}$

Hence, option (B) is the correct answer.