OnlineCBT
Mathematics · JEE | DIFFERENTIAL EQUATIONS
हिन्दीWhat is 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 \( \frac{2x}{y^2} \)?
View Answer & Solution
Correct answer · Option B
\( y^3 = 3x^2 + 35 \)
Explanation
Setting \( \frac{dy}{dx} = \frac{2x}{y^2} \) and separating variables gives \( y^2 dy = 2x dx \). Integrating yields \( \frac{y^3}{3} = x^2 + C \). Using \( (-2, 3) \), \( \frac{27}{3} = 4 + C \implies C = 5 \), so \( y^3 = 3x^2 + 15 \)... wait, \( 3 = 9 \), hence \( y^3 = 3x^2 + 15 \).
