An angle between the curves x^2 y = 1 and y(x | vedclass

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

Question

(D) Given curves are $C_1: x^2 y = 1$ and $C_2: y(x^2 + 1) = 2$. First,find the intersection points by substituting $y = 1/x^2$ into the second equati

Vedclass · Accepted Answer

$\operatorname{Tan}^{-1} \frac{1}{3}$

Vedclass · Answer

(D) Given curves are $C_1: x^2 y = 1$ and $C_2: y(x^2 + 1) = 2$. First,find the intersection points by substituting $y = 1/x^2$ into the second equation: $(1/x^2)(x^2 + 1) = 2 \implies 1 + 1/x^2 = 2 \implies 1/x^2 = 1 \implies x^2 = 1 \implies x = \pm 1$. For $x = 1$,$y = 1$. For $x = -1$,$y = 1$. Intersection points are $(1, 1)$ and $(-1, 1)$. For $C_1$,$y = x^{-2} \implies dy/dx = -2x^{-3} = -2/x^3$. At $(1, 1)$,$m_1 = -2$. For $C_2$,$y = 2/(x^2 + 1) \implies dy/dx = -2(2x)/(x^2 + 1)^2 = -4x/(x^2 + 1)^2$. At $(1, 1)$,$m_2 = -4(1)/(1+1)^2 = -4/4 = -1$. The angle $	heta$ between the curves is given by $	an 	heta = |(m_1 - m_2) / (1 + m_1 m_2)|$. $	an 	heta = |(-2 - (-1)) / (1 + (-2)(-1))| = |-1 / (1 + 2)| = |-1/3| = 1/3$. Thus,$	heta = \operatorname{Tan}^{-1}(1/3)$.

An angle between the curves $x^2 y = 1$ and $y(x^2 + 1) = 2$ is

Similar Questions

The sum of the intercepts on the axes cut off by the tangent to the curve $x^{2/3} + y^{2/3} = 2$ at the point $(1, 1)$ is:

The tangent drawn at the point $(0, 1)$ on the curve $y = e^{2x}$ meets the $x-$axis at the point:

The sum of the intercepts made by a tangent drawn to the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$ at the point $(a, b)$ on the coordinate axes is:

The equation of a normal to the curve $x=4 \sec \theta$ and $y=4 \tan^2 \theta$ at $\theta=\frac{\pi}{4}$ is

For any point on a curve,what is $\sqrt{\frac{\text{subnormal}}{\text{subtangent}}}$ equal to?

Vedclass Test Series

Exam Paper Generator

Online Exam Module