Question

For the curve $y^{n} = a^{n - 1}x,$ the sub-normal at any point is

constant, the value of n must be

• A. 2
• B. 3
• C. 0
• D. 1

Answer 1

Answer: (A)

2

Answer 2

$y^{n} = a^{n - 1}x$ ⇒ $ny^{n - 1}\frac{dy}{dx} = a^{n - 1}$ ⇒ $\left( \frac{dy}{dx} \right) = \frac{a^{n - 1}}{ny^{n - 1}}$

$\therefore$ Length of the subnormal = $y\frac{dy}{dx} = \frac{ya^{n - 1}}{ny^{n - 1}} = \frac{a^{n - 1}y^{2 - n}}{n}$

We also know that if the subnormal is constant, then $\frac{a^{n - 1}}{n}.y^{2 - n}$ should not contain y.

Therefore, $2 - n = 0$ or $n = 2$.