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

Concept: 

A slope of the tangent to a curve at any point (x, y) is given by \(\rm\frac{\mathrm{d} y}{\mathrm{d} x}\) . 

Calculation: 

We know that slope of the tangent, 

 \(\rm\frac{\mathrm{d} y}{\mathrm{d} x}\) = \(\rm \frac{3x}{y^{2}}\) 

⇒ y² dy = 3x dx 

On integrating both sides , we get

\(\rm \int y^{2}dy = \int 3x \ dx\)

⇒ \(\rm \frac{y^{3}}{3} = \frac{3}{2}x^{2} +C\)          .... (i) 

Given that curve passes through point ( -2, 3 ) ,Thus, substituting x = -2 and y = 3 in (i), we get 

\(\rm \frac{27}{3} = \frac{12}{2} + C\)

 ⇒ C = 3 

Putting C = 3 in (i), we get 

\(\rm \frac{y^{3}}{3} = \frac{3}{2}x^{2} +3\)  is 

The required equation of curve  .

The correct option is 2.