The equation of the curve passing through the | vedclass

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(B) Given that the product of the slope of the tangent $\frac{dy}{dx}$ and the $y$-coordinate is equal to the $x$-coordinate,we have the differential

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$y^2 - x^2 = 4$

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(B) Given that the product of the slope of the tangent $\frac{dy}{dx}$ and the $y$-coordinate is equal to the $x$-coordinate,we have the differential equation: $\frac{dy}{dx} \cdot y = x$ Separating the variables,we get: $y \, dy = x \, dx$ Integrating both sides: $\int y \, dy = \int x \, dx$ $\frac{y^2}{2} = \frac{x^2}{2} + C$ The curve passes through the point $(0, -2)$. Substituting $x = 0$ and $y = -2$: $\frac{(-2)^2}{2} = \frac{0^2}{2} + C$ $\frac{4}{2} = 0 + C \Rightarrow C = 2$ Substituting $C = 2$ back into the equation: $\frac{y^2}{2} = \frac{x^2}{2} + 2$ Multiplying by $2$: $y^2 = x^2 + 4$ $y^2 - x^2 = 4$

The equation of the curve passing through the point $(0, -2)$ given that at any point $(x, y)$ on the curve,the product of the slope of its tangent and the $y$-coordinate of the point is equal to the $x$-coordinate of the point,is

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