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