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(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{2x}{y^2}$.Separating the variables,we get $y^2 dy =

Vedclass · Accepted Answer

$\frac{y^3}{3} = x^2 + 5$

Vedclass · Answer

(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{2x}{y^2}$.Separating the variables,we get $y^2 dy = 2x dx$.Integrating both sides,we have $\int y^2 dy = \int 2x dx$.This gives $\frac{y^3}{3} = x^2 + C$.Since the curve passes through the point $(-2, 3)$,we substitute $x = -2$ and $y = 3$ into the equation:$\frac{3^3}{3} = (-2)^2 + C$$\frac{27}{3} = 4 + C$$9 = 4 + C$$C = 5$.Substituting $C = 5$ back into the equation,we get $\frac{y^3}{3} = x^2 + 5$.

Find the equation of a curve passing through the point $(-2, 3),$ given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$.

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