The orthogonal trajectories of the family of curves $a^{n-1} y=x^n$ are given by ( $a$ is the arbitrary constant)
Answer & explanation
Correct answer: option 2
The equation of the given family of curves is
$a^{n-1} y=x^n$ .......(i)
$\Rightarrow (n-1) \log a+\log y=n \log x$
Differentiating w.r.t. $x$, we get
$\frac{1}{y} \frac{d y}{d x}=\frac{n}{x}$ ......(ii)
This is the differential equation of the family of curves given in (i).
The differential equation of the orthogonal trajectories of (i) is obtained by replacing $\frac{d y}{d x} b y-\frac{d x}{d y}$ in (ii).
Replacing $\frac{d y}{d x} b y-\frac{d x}{d y}$ in (ii), we get
$\frac{1}{y} \times-\frac{d x}{d y}=\frac{n}{x} \Rightarrow x d x+n y d y=0$
On integrating, we get
$\frac{x^2}{2}+n \frac{y^2}{2}=C \Rightarrow x^2+n y^2=2 C$