The orthogonal trajectories of the family of | vedclass

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(B) Given the curve $x^{2/3} + y^{2/3} = a^{2/3}$.Differentiating with respect to $x$,we get:$\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx}

Vedclass · Accepted Answer

$x^{4/3} - y^{4/3} = c$

Vedclass · Answer

(B) Given the curve $x^{2/3} + y^{2/3} = a^{2/3}$.Differentiating with respect to $x$,we get:$\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$$\Rightarrow x^{-1/3} + y^{-1/3} \frac{dy}{dx} = 0$$\Rightarrow \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3}$.For orthogonal trajectories,replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$:$-\frac{dx}{dy} = -\left(\frac{y}{x}\right)^{1/3}$$\Rightarrow \frac{dx}{dy} = \frac{y^{1/3}}{x^{1/3}}$$\Rightarrow x^{1/3} dx = y^{1/3} dy$.Integrating both sides:$\int x^{1/3} dx = \int y^{1/3} dy$$\Rightarrow \frac{x^{4/3}}{4/3} = \frac{y^{4/3}}{4/3} + C$$\Rightarrow x^{4/3} - y^{4/3} = C'$ (where $C' = \frac{4}{3}C$ is a constant).Thus,the correct option is $B$.

The orthogonal trajectories of the family of curves $x^{2/3} + y^{2/3} = a^{2/3}$,where $a$ is an arbitrary constant,are given by:

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