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

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(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

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Hyperbola

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(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

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