Question

Orthogonal trajectories of family of the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}},$ where $a$ is any arbitrary constant, is:
(where $c$ is integration constant)

• A. $x^{\frac{2}{3}}-y^{\frac{2}{3}}=c$
• B. $x^{\frac{4}{3}}-y^{\frac{4}{3}}=c$
• C. $x^{\frac{4}{3}}+y^{\frac{4}{3}}=c$
• D. $x^{\frac{1}{3}}-y^{\frac{1}{3}}=c$

Answer 1

The correct option is B $x^{\frac{4}{3}}-y^{\frac{4}{3}}=c$

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
$\Rightarrow \frac{2}{3}{x}^{-\frac{1}{3}}+\frac{2}{3}{y}^{-\frac{1}{3}}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$
In order to get orthogonal trajectory, replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$
$\Rightarrow -\frac{dx}{dy}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$
$\Rightarrow \int x^{\frac{1}{3}}dx=\int y^{\frac{1}{3}}dy$
$\Rightarrow x^{\frac{4}{3}}-y^{\frac{4}{3}}=c$