If the slope of the tangent to the curve \(y = y(x)\) at any point \((x, y)\) is \(\displaystyle\frac{2x}{y^2}\) and the curve passes through the point \(\left(\displaystyle\frac{1}{\sqrt{3}}, 1\right)\), then equation of curve is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(y^3 = 3x^2\)
The slope of the tangent is \(\frac{dy}{dx}\), so
\( \displaystyle \frac{dy}{dx} = \frac{2x}{y^2} \)
The variables separate:
\(y^2\,dy = 2x\,dx\)
\( \displaystyle \int y^2\,dy = \int 2x\,dx \)
\( \displaystyle \frac{y^3}{3} = x^2 + C \)
The curve passes through \(\left(\frac{1}{\sqrt{3}}, 1\right)\):
\( \displaystyle \frac{1}{3} = \frac{1}{3} + C \)
\(C = 0\)
So \(\frac{y^3}{3} = x^2\), that is \(y^3 = 3x^2\).