Any curve passes through the point (3, -4). I | vedclass

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

Vedclass · Accepted Answer

$y = \frac{-4}{9}x^2$

Vedclass · Answer

(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{2y}{x}$.Separating the variables,we get $\frac{dy}{y} = \frac{2dx}{x}$.Integrating both sides,we have $\int \frac{dy}{y} = 2 \int \frac{dx}{x}$,which gives $\ln|y| = 2 \ln|x| + C$.This simplifies to $\ln|y| = \ln|x^2| + C$,or $y = kx^2$,where $k = e^C$.Since the curve passes through $(3, -4)$,we substitute these values: $-4 = k(3)^2$,which implies $-4 = 9k$,so $k = -\frac{4}{9}$.Thus,the equation of the curve is $y = -\frac{4}{9}x^2$,which can be rewritten as $4x^2 + 9y = 0$.

Any curve passes through the point $(3, -4)$. If the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2y}{x}$,then the equation of the curve is . . . . . . .

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