Find the equation of a curve passing through | vedclass

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Question

(A) Let the slope of the tangent at any point $(x, y)$ be $\frac{dy}{dx}$.According to the problem,the product of the slope and the $y$-coordinate is

Vedclass · Accepted Answer

$y^{2} - x^{2} = 4$

Vedclass · Answer

(A) Let the slope of the tangent at any point $(x, y)$ be $\frac{dy}{dx}$.According to the problem,the product of the slope and the $y$-coordinate is equal to the $x$-coordinate:$y \frac{dy}{dx} = x$Separating the variables,we get:$y \, dy = x \, dx$Integrating both sides:$\int y \, dy = \int x \, dx$$\frac{y^{2}}{2} = \frac{x^{2}}{2} + C$$y^{2} - x^{2} = 2C$Let $2C = K$,so $y^{2} - x^{2} = K$.The curve passes through $(0, -2)$,so substitute $x = 0$ and $y = -2$:$(-2)^{2} - (0)^{2} = K$$4 = K$Thus,the equation of the curve is $y^{2} - x^{2} = 4$.

Find the equation of a 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.

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