Let C_{1} be the curve obtained by the soluti | vedclass

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

Question

(B) For curve $C_{1}$: $\frac{dy}{dx} = \frac{y^{2} - x^{2}}{2xy}$.Put $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.$v + x\frac{dv}{dx} = \frac{

Vedclass · Accepted Answer

$\frac{\pi}{2} - 1$

Vedclass · Answer

(B) For curve $C_{1}$: $\frac{dy}{dx} = \frac{y^{2} - x^{2}}{2xy}$.Put $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.$v + x\frac{dv}{dx} = \frac{v^{2}x^{2} - x^{2}}{2x(vx)} = \frac{v^{2} - 1}{2v}$.$x\frac{dv}{dx} = \frac{v^{2} - 1}{2v} - v = \frac{v^{2} - 1 - 2v^{2}}{2v} = \frac{-(v^{2} + 1)}{2v}$.Separating variables: $\frac{2v}{v^{2} + 1} dv = -\frac{dx}{x}$.Integrating both sides: $\ln(v^{2} + 1) = -\ln x + C$.$\ln(\frac{y^{2}}{x^{2}} + 1) = -\ln x + C \Rightarrow \ln(\frac{x^{2} + y^{2}}{x^{2}}) + \ln x = C \Rightarrow \ln(\frac{x^{2} + y^{2}}{x}) = C$.Since it passes through $(1, 1)$,$\ln(\frac{1+1}{1}) = C \Rightarrow C = \ln 2$.So,$\frac{x^{2} + y^{2}}{x} = 2 \Rightarrow x^{2} + y^{2} - 2x = 0$ (Circle with center $(1, 0)$ and radius $1$).For curve $C_{2}$: $\frac{dy}{dx} = \frac{2xy}{x^{2} - y^{2}}$.This is the reciprocal of the slope of $C_{1}$ with a sign change,implying orthogonality or symmetry. Solving similarly with $y=vx$ leads to $x^{2} + y^{2} - 2y = 0$ (Circle with center $(0, 1)$ and radius $1$).The area enclosed by these two circles is $2 \int_{0}^{1} (\sqrt{1 - (x-1)^{2}} - x) dx = 2 [\frac{\pi}{4} - \frac{1}{2}] = \frac{\pi}{2} - 1$.

Similar Questions

Area of the region (in square units) enclosed by the curves $y^2=8(x+2)$,$y^2=4(1-x)$ and the $Y$-axis is

The area bounded between the curve $x^{2}=y$ and the line $y=4x$ is

The area (in sq. units) of the region described by $\{(x, y) : y^2 \leq 2x \text{ and } y \geq 4x - 1\}$ is

The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 36$ which is outside the parabola $y^{2} = 9x$ is:

One of the points of intersection of the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(\ell \sqrt{5}+m)-n \log_{e}(1+\sqrt{5})$,where $\ell, m, n \in N$. Then $\ell+m+n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module