Question

The orthogonal trajectories of the family of curves $a^{n-1}y=x^n$ are given by:

• A. $x^n+n^{2}y=$const
• B. $n{y}^2+x^2=$const
• C. $n^{2}x+y^n=$const
• D. $n^{2x}-y^n=$const

Answer 1

Answer: (B)

$n{y}^2+x^2=$const

$a^{n-1}y=x^n$

Differentiating, we get

$a^{n-1}\frac{dy}{dx}=n{x}^{n-1}\Rightarrow a^{n-1}=n{x}^{n-1}\frac{dx}{dy}$

Putting this value in the given equation, we have

$n{x}^{n-1}\frac{dx}{dy}y=x^n$

Replacing $\frac{dy}{dx}$ by $-\frac{dx}{dy}$, we have

$ny=-\frac{dx}{dy}\Rightarrow nydy+xdx=0\Rightarrow n{y}^2+x^2=c$