Question

The orthogonal trajectories of the family of curves $x^{2/3}+y^{2/3}=a^{2/3}$, where $a$ is the parameter, is

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

Answer 1

The correct option is C $x^{4/3}-y^{4/3}=c^{2/3}$

The given family of curves $x^{2/3}+y^{2/3}=a^{2/3}$
Differentiating w.r.t. $x,$ we get
$\frac{2}{3}{x}^{-1/3}+\frac{2}{3}{y}^{-1/3}{y}^{\prime }=0$
$\Rightarrow \frac{dy}{dx}=-\frac{y^{1/3}}{x^{1/3}}$
Replacing $\frac{dy}{dx}$ by $-\frac{dx}{dy},$ we get
$-\frac{dx}{dy}=-\frac{y^{1/3}}{x^{1/3}}$
$\Rightarrow x^{1/3}dx=y^{1/3}dy$
$\Rightarrow \frac{3}{4}{x}^{4/3}=\frac{3}{4}{y}^{4/3}+k$
$\Rightarrow x^{4/3}-y^{4/3}=\frac{4}{3}k$
$\therefore x^{4/3}-y^{4/3}=c^{2/3}(c^{2/3}=\frac{4}{3}k)$