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 the point on the curve where the tangent is vertical, then H and V are respectively given by:

1. $\\{ (0,0)\\} ,\\{ (0,1)\\}$
2. $\Phi ,\\{ (1,1)\\}$
3. $\\{ (1,1)\\} ,\\{ (0,0)\\}$
4. None of these

Answer 1

Hint : Here, we are given an equation of the curve. We will apply derivatives on both sides with respect to x. We need to find the values of H and V. At the points where the tangent is horizontal, the slope of the tangent is 0 and when the tangent is vertical, then the slope of the tangent is infinity. Using this slope value we can find the value of y and substitute in the given curve equation to find the final output.

Complete step-by-step answer :
Given the curve as,
${y^3} - 3xy + 2 = 0$
Apply derivate on both the sides with respect to x, we will get,
$\Rightarrow 3{y^2}\dfrac{{dy}}{{dx}} - 3x\dfrac{{dy}}{{dx}} - 3y = 0$
$\Rightarrow \dfrac{{dy}}{{dx}}(3{y^2} - 3x) = 3y$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{3y}}{{3{y^2} - 3x}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{{{y^2} - x}}$
First,
For the points where the tangent is horizontal.
This means, the slope of the tangent is 0.
$\therefore \dfrac{{dy}}{{dx}} = 0$
$\Rightarrow \dfrac{y}{{{y^2} - x}} = 0$
$\Rightarrow y = 0$
Substituting this value of y in given equation of curve, we will get,
$\Rightarrow 0 - 3x(0) + 2 = 0$
$\Rightarrow 2 = 0$ , which is not possible.
Thus, y = 0 does not satisfy the given equation of the curve therefore y cannot lie on the curve.
$\therefore H = \\{ \Phi \\} = \Phi$.
Next,
For the points where tangent is vertical, then
This means, the slope of the tangent is infinity.
$\therefore \dfrac{{dy}}{{dx}} = \infty$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{0}$
$\Rightarrow \dfrac{y}{{{y^2} - x}} = \dfrac{1}{0}$
$\Rightarrow {y^2} - x = 0$
$\Rightarrow {y^2} = x$
$\Rightarrow x = {y^2}$
Substituting this value of x in given equation of curve, we will get,
$\Rightarrow {y^3} - 3({y^2})y + 2 = 0$
$\Rightarrow {y^3} - 3{y^3} + 2 = 0$
$\Rightarrow - 2{y^3} + 2 = 0$
$\Rightarrow 2{y^3} - 2 = 0$
$\Rightarrow {y^3} - 1 = 0$
$\Rightarrow {y^3} = 1$
$\Rightarrow y = 1$
When y=1, then x=1.
$\therefore V = \\{ (1,1)\\}$
Hence, for a given curve ${y^3} - 3xy + 2 = 0$, the value of the tangent when its horizontal is a null set and at vertical it is {(1,1)}.
So, the correct answer is “Option 2”.

Note : A tangent is a line which represents the slope of a curve at that point. A slope of a line is calculated by dividing the change in height by the change in horizontal distance. Tangent is a straight line (or smooth curve) that touches a given curve at one point and at that point the slope of the curve is equal to that of the tangent.