If A = {P(alpha, beta) mid text{the tangent d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given curve: $y^3 - 3xy + 2 = 0$. Differentiating with respect to $x$: $3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0$. $\frac{dy}{dx} (3y^2 - 3x

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given curve: $y^3 - 3xy + 2 = 0$. Differentiating with respect to $x$: $3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0$. $\frac{dy}{dx} (3y^2 - 3x) = 3y \implies \frac{dy}{dx} = \frac{y}{y^2 - x}$. For a horizontal tangent, $\frac{dy}{dx} = 0 \implies y = 0$. Substituting $y = 0$ into the curve equation: $0^3 - 3x(0) + 2 = 0 \implies 2 = 0$, which is impossible. Thus, $n(A) = 0$. For a vertical tangent, $\frac{dy}{dx} = \infty \implies y^2 - x = 0 \implies x = y^2$. Substituting $x = y^2$ into the curve equation: $y^3 - 3(y^2)y + 2 = 0 \implies y^3 - 3y^3 + 2 = 0 \implies -2y^3 = -2 \implies y^3 = 1 \implies y = 1$. If $y = 1$, then $x = 1^2 = 1$. The point is $(1, 1)$. Thus, $n(B) = 1$. Therefore, $n(A) + n(B) = 0 + 1 = 1$.

If $A = \{P(\alpha, \beta) \mid \text{the tangent drawn at } P \text{ to the curve } y^3 - 3xy + 2 = 0 \text{ is a horizontal line}\}$ and $B = \{Q(a, b) \mid \text{the tangent drawn at } Q \text{ to the curve } y^3 - 3xy + 2 = 0 \text{ is a vertical line}\}$, then $n(A) + n(B) = $

Similar Questions

The angle between the curves $y=\sin x$ and $y=\cos x$ is

What is the angle between the curves $y = x^2$ and $x = y^2$ at the point $(1, 1)$?

The equation of the tangent to the curve $x y^5+2 x^2 y-x^3+y+1=0$ at $x=0$ is

Show that the line $\frac{x}{a} + \frac{y}{b} = 1$ touches the curve $y = b \cdot e^{-x / a}$ at the point where the curve intersects the $y$-axis.

If the normal to the curve $x^{2/3} + y^{2/3} = a^{2/3}$ makes an angle $\phi$ with the $X$-axis,then the equation of that normal is

Vedclass Test Series

Exam Paper Generator

Online Exam Module