The curve passing through the point (1,2) giv | vedclass

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

Question

(D) Given the slope of the tangent $\frac{dy}{dx} = \frac{3x}{y}$.By separating the variables,we get $y \, dy = 3x \, dx$.Integrating both sides,we ha

Vedclass · Accepted Answer

Hyperbola

Vedclass · Answer

(D) Given the slope of the tangent $\frac{dy}{dx} = \frac{3x}{y}$.By separating the variables,we get $y \, dy = 3x \, dx$.Integrating both sides,we have $\int y \, dy = \int 3x \, dx$,which gives $\frac{y^2}{2} = \frac{3x^2}{2} + C$.Multiplying by $2$,we get $y^2 = 3x^2 + 2C$,or $y^2 - 3x^2 = K$ where $K = 2C$.Since the curve passes through $(1, 2)$,we substitute $x=1$ and $y=2$ into the equation: $2^2 - 3(1)^2 = K$,so $4 - 3 = K$,which means $K = 1$.The equation of the curve is $y^2 - 3x^2 = 1$.This equation is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,which represents a Hyperbola.

The curve passing through the point $(1,2)$ given that the slope of the tangent at any point $(x, y)$ is $\frac{3x}{y}$ represents

Similar Questions

The particular solution of the differential equation $\cos \left(\frac{dy}{dx}\right) = 0.5$ with the condition $y = 1$ at $x = 0$ is:

The general solution of $\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y$ is

The general solution of the differential equation $\frac{dy}{dx} = 1 + x + y + xy$ is

The general solution of the differential equation $\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0$ is (where $C$ is a constant of integration)

Find the general solution of the differential equation: $\frac{dy}{dx} = \sqrt{4 - y^2}$ where $-2 < y < 2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module