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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 $a^2$, then $n=$
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The correct answer is:
$\frac{3}{2}$
We have,
$y=\frac{x^n}{a^{n-1}} \quad \therefore \frac{d y}{d x}=\frac{n x^{n-1}}{a^{n-1}}$
$\therefore$ 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 $a^2$
$\therefore \quad 2 n-1=2 \quad \Rightarrow \quad n=\frac{3}{2}$
$y=\frac{x^n}{a^{n-1}} \quad \therefore \frac{d y}{d x}=\frac{n x^{n-1}}{a^{n-1}}$
$\therefore$ 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 $a^2$
$\therefore \quad 2 n-1=2 \quad \Rightarrow \quad n=\frac{3}{2}$
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