Find the equation of a curve which passes through the point (−2, 3)

1.	Find the equation of a curve which passes through the point (−2, 3) and the slope of whose tangent at any point (x, y) is \(\frac{2x}{y^2}.\)
Answer» Given that the slope of tangent to the curve at the point (x, y) is \(\frac{2x}{y^2}\) Therefore, \(\frac{dy}{dx} = \frac{2x}{y^2}\) ⇒ \(y^2 dy = 2x dx.\) ⇒ ∫ y²y = ∫ 2x dx (By integrating both sides of above equation) ⇒ \(\frac{y^3}{3}\) = 2\(\frac{x^2}{2}\) + C, where C is an integral constant. \(\left(\because \int x^n dx = \frac{x^{n+1}}{n+1}\right)\) ⇒ y^3 = 3x^2 + 3C which is equation of given curve whose slope of tangent at any point (x, y) is \(\frac{2x}{y^2}\) Since, the given curve is passing through the point (−2, 3). Hence, satisfies its equation, therefore, by putting x = −2, y = 3, we get 3³= 3(−2)² + 3C ⇒ 3C = 27 − 12 = 15. Therefore, the equation of the curve is y³ = 3x² + 15. (By putting 3C = 15 in equation of curve) Hence, the equation of a curve whose slope of tangent at any point (x, y) and which passes through the point (-2,3) is y³ = 3x² + 15.

Find the equation of a curve which passes through the point (−2, 3) and the slope of whose tangent at any point (x, y) is \(\frac{2x}{y^2}.\)

Answer»

Given that the slope of tangent to the curve at the point (x, y) is \(\frac{2x}{y^2}\)

Therefore, \(\frac{dy}{dx} = \frac{2x}{y^2}\) ⇒ \(y^2 dy = 2x dx.\)

⇒ ∫ y²y = ∫ 2x dx (By integrating both sides of above equation)

⇒ \(\frac{y^3}{3}\) = 2\(\frac{x^2}{2}\) + C, where C is an integral constant. \(\left(\because \int x^n dx = \frac{x^{n+1}}{n+1}\right)\)

⇒ y^3 = 3x^2 + 3C which is equation of given curve whose slope of tangent at any point (x, y) is \(\frac{2x}{y^2}\)

Since, the given curve is passing through the point (−2, 3). Hence, satisfies its equation, therefore, by putting x = −2, y = 3, we get 3³= 3(−2)² + 3C

⇒ 3C = 27 − 12 = 15. Therefore, the equation of the curve is

y³ = 3x² + 15. (By putting 3C = 15 in equation of curve)

Hence, the equation of a curve whose slope of tangent at any point (x, y) and which passes through the point (-2,3) is y³ = 3x² + 15.

Find the equation of a curve which passes through the point (−2, 3) and the slope of whose tangent at any point (x, y) is \(\frac{2x}{y^2}.\)

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