The angle between the tangents drawn at (0,0) | vedclass

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

Question

(D) For the curve $y^3-x^2 y+5 y-2 x=0$,differentiating with respect to $x$ gives: $3 y^2 \frac{dy}{dx} - (x^2 \frac{dy}{dx} + 2xy) + 5 \frac{dy}{dx}

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) For the curve $y^3-x^2 y+5 y-2 x=0$,differentiating with respect to $x$ gives: $3 y^2 \frac{dy}{dx} - (x^2 \frac{dy}{dx} + 2xy) + 5 \frac{dy}{dx} - 2 = 0$ At $(0,0)$,this simplifies to $5 \frac{dy}{dx} - 2 = 0$,so the slope $m_1 = \frac{2}{5}$. For the curve $x^4-x^3 y^2+5 x+2 y=0$,differentiating with respect to $x$ gives: $4x^3 - (x^3 \cdot 2y \frac{dy}{dx} + 3x^2 y^2) + 5 + 2 \frac{dy}{dx} = 0$ At $(0,0)$,this simplifies to $5 + 2 \frac{dy}{dx} = 0$,so the slope $m_2 = -\frac{5}{2}$. Since $m_1 	imes m_2 = (\frac{2}{5}) 	imes (-\frac{5}{2}) = -1$,the tangents are perpendicular. Therefore,the angle between them is $\frac{\pi}{2}$.

The angle between the tangents drawn at $(0,0)$ to the curves $y^3-x^2 y+5 y-2 x=0$ and $x^4-x^3 y^2+5 x+2 y=0$ is

Similar Questions

The sum of the lengths of the subtangent and the subnormal drawn at $\theta = \frac{\pi}{3}$ on the cycloid $x = a(\theta - \sin \theta)$,$y = a(1 - \cos \theta)$ is

If the normal to the curve $y = f(x)$ is parallel to the $x-$axis,then the correct statement is:

The equation of the normal to the curve $y = \sin \left( \frac{\pi x}{2} \right)$ at $(1, 1)$ is

Let $m$ be the slope of the normal $L$ drawn at $(1, 2)$ to the curve $x = t^2 - 7t + 7, y = t^2 - 4t - 10$ and $ax + by + c = 0$ be the equation of the normal $L$. If the $G$.$C$.$D$. of $(a, b, c)$ is $1$, then $m(a + b + c) =$

The chord of the curve $y=x^{2}+2ax+b$ joining the points where $x=\alpha$ and $x=\beta$ is parallel to the tangent to the curve at abscissa $x$ equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module