Constant subnormal

Question

Find the value of $n$ so that the subnormal at any point on the curve $xy^n = a^{n + 1}$ may be constant.

I tried and found that slope of normal will be $nx/y$

But how to proceed ?

Answer 1

The length of the subnormal for a given curve $P(x,y)$ is given by $$\text{Length of subnormal} = y\tan \phi = y\frac{dy}{dx}$$ where $\phi$ is the angle of inclination of the tangent with respect to the x-axis.

For our curve, $xy^n = a^{n+1}$, we get the value of $y\frac{dy}{dx}$ as $ -\frac{y^2}{nx}$.

What can we conclude from this?

We know that the subnormal for the parabola is a constant. You can see here for a proof. Thus, the equation of the curve $xy^n = a^{n+1}$ represents a parabola. Thus, on rearranging, we get, $$y^{-n} = a^{-(n+1)}x$$ This gives us $n =-2$. Hope it helps.

Answer 2

Differentiating both sides of the given equation:

$y^n + n x y^{n-1}\dfrac{dy}{dx}= 0 $

$\implies \dfrac{dy}{dx}= -\dfrac{y^n\times y}{nxy^{n-1}\times y}= -\dfrac{y^{n+1}}{na^{n+1}} $

Length of subnormal is given by $y y' = -\dfrac{y^{n+2}}{na^{n+1}}$, which is constant for $n =-2$