Application of Derivatives 1 Question 16
####16. 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 horizontal and $V$ is the set of points on the curve $C$, where the tangent is vertical, then $H=\ldots$ and $V=\ldots$.
(1994, 2M)
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Answer:
($H= \theta , V= { 1,1}$)
Solution:
- Given, $y^{3}-3 x y+2=0$
On differentiating w.r.t. $x$, we get
$ \begin{aligned} 3 y^{2} \frac{d y}{d x}-3 x \frac{d y}{d x}-3 y & =0 \\ \Rightarrow \quad \frac{d y}{d x}\left(3 y^{2}-3 x\right) & =3 y \quad \Rightarrow \quad \frac{d y}{d x}=\frac{3 y}{3 y^{2}-3 x} \end{aligned} $
For the points where tangent is horizontal, the slope of tangent is zero.
i.e.
$ \frac{d y}{d x}=0 \Rightarrow \frac{3 y}{3 y^{2}-3 x}=0 $
$\Rightarrow y=0$ but $y=0$ does not satisfy the given equation of the curve, therefore $y$ cannot lie on the curve.
So, $\quad H=\varphi \quad$ [null set]
For the points where tangent is vertical, $\frac{d y}{d x}=\infty$
$ \begin{array}{rlrl} \Rightarrow & & \frac{y}{y^{2}-x} & =\infty \\ \Rightarrow & y^{2}-x & =0 \\ \Rightarrow & y^{2} & =x \end{array} $
On putting this value in the given equation of the curve, we get
$ \begin{aligned} & y^{3}-3 \cdot y^{2} \cdot y+2=0 \\ & \Rightarrow \quad-2 y^{3}+2=0 \\ & \Rightarrow \quad y^{3}-1=0 \Rightarrow y^{3}=1 \\ & \Rightarrow \quad y=1, x=1 \\ & V={1,1} \end{aligned} $
