Question

Let $a$ be a fixed positive real number and $n$ be an arbitrary constant. For the curve $y=\frac{x^{n}}{a^{n-1}}$, if the length of the subnormal at any point $(\alpha, \beta)$ is proportional to $\alpha^{2}$, the $n=$

• A. 2
• B. 1
• C. 0
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

We have,
$y=\frac{x^{n}}{a^{n-1}} $
$ \therefore \frac{d y}{d x}=\frac{n x^{n-1}}{a^{n-1}}$
$\therefore $ Length of subnormal
$=y \frac{d y}{d x}=\frac{x^{n}}{a^{n-1}} \times n \frac{x^{n-1}}{a^{n-1}}$ $=n \frac{x^{2 n-1}}{a^{2 n-2}}$
$\therefore $ Length of subnormal at
$(\alpha, \beta)=\frac{n \alpha}{a^{2 n-2}}$
Now, it is given that
Length of subnormal is proportional to $a^{2}$
$\therefore 2 n-1=2$
$ \Rightarrow n=\frac{3}{2}$