Question

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

Answer 1

The slope of the tangent to a curve is given by $\frac{dy}{dx}$.

Therefore,

$\frac{dy}{dx}=\frac{2x}{y^2}$

$y^{2}dy=2xdx$

Integrating both sides,

$\int y^{2}dy=\int 2xdx$

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

When, $x=-2,y=3$

Then,

$9=4+C$

$C=5$

Therefore, the equation of the required curve is $\frac{y^3}{3}=x^2+5$.