Find orthogonal trajectories of the parabola family ay²=x³

1.	Find orthogonal trajectories of the parabola family ay²=x³
Answer» Working Rule :- 1. Differentiate the given curve F(x, y, a) 2. OBTAIN the value of arbitrary constant a. 3. Substituting the value of arbitrary constant 'a' in f(x, y, a) to eliminate a. 4. Then replace dy/dx by - dx/dy. 5. If we get the same equation back, curve is CALLED Self - Orthogonal otherwise integrating both sides. Let's solve the problem now!!!! $\green{\large\underline{\bf{Solution-}}}$ The given equation is $\bf :\longmapsto\: {ay}^{2} = {x}^{3} - - - (1)$ Differentiating both sides W. r. t. x, we get $\rm :\longmapsto\:\dfrac{d}{dx} {ay}^{2} = \dfrac{d}{dx} {x}^{3}$ $\rm :\longmapsto\: \: 2ay\dfrac{dy}{dx} = 3 {x}^{2}$ $\bf\implies \:a = \dfrac{3 {x}^{2} }{2y\dfrac{dy}{dx} } - - - (2)$ On substituting the value of 'a' in equation (1), we get $\rm :\longmapsto\:\dfrac{3 {x}^{2} }{2y\dfrac{dy}{dx}} {y}^{2} = {x}^{3}$ $\rm :\longmapsto\: {3x}^{2}y\dfrac{dx}{dy} = 2{x}^{3}$ $\rm :\longmapsto\: 3y\dfrac{dx}{dy} = 2x$ $\red{\bf :\longmapsto\:On \: replacing \: \dfrac{dx}{dy} \: by \: - \dfrac{dy}{dx}, \: we \: get}$ $\rm :\longmapsto\: - \: 3y\dfrac{dy}{dx} = 2x$ $\rm :\longmapsto\: - 3ydy = 2xdx$ On integrating both sides, we get $\displaystyle \: \rm :\longmapsto\: - \int 3ydy = \int 2xdx$ $\rm :\longmapsto\: - \dfrac{3 {y}^{2} }{2} = {x}^{2} + c$ $\rm :\longmapsto\: - 3 {y}^{2} = 2{x}^{2} + 2c$ $\rm :\longmapsto\: 3 {y}^{2} + 2{x}^{2} + 2c = 0$ $\rm :\longmapsto\: 3 {y}^{2} + 2{x}^{2} +d = 0 \: \: \: \: where \: d \: = \: 2c$ Thus, $\: \: \: \underbrace{\boxed{ \bf \: {2x}^{2} + {3y}^{2} + d = 0 \: is \: required \: equation}}$ Basic Formula's Used :- $\boxed{ \sf \: \dfrac{d}{dx} {x}^{2} = 2x}$ $\boxed{ \sf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}$ $\boxed{ \sf \: \int \: {x}^{n}dx = \dfrac{ {x}^{n + 1} }{n + 1} + c}$

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