If the area of the triangle included between the axes and any tangent to the curve $x^{n} y = a^{n}$ is constant, then the value of $n$ is

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Solution

Any point on the curve $x^{n} y = a^{n}$ is,
$P (a t, \frac{1}{t^{n}})$

Differentiate w.r.t $x$ , we get
$n x^{n - 1} y + x^{n} \frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d x} = - n \frac{y}{x}$
$\Rightarrow \frac{d y}{d x} = - \frac{n}{a t^{n + 1}}$

The equation of the tangent at $P (a t, \frac{1}{t^{n}})$ is
$y - \frac{1}{t^{n}} = - \frac{n}{a t^{n + 1}} (x - a t)$
This meets the co-orinate axes at
$A (\frac{a t (n + 1)}{n}, 0) B (0, \frac{(n + 1)}{t^{n}})$

$Area of Δ A O B = \frac{1}{2} (O A \times O B) = \frac{1}{2} (\frac{a t (n + 1)}{n}) (\frac{(n + 1)}{t^{n}}) = \frac{a (n + 1)^{2}}{2 n} t^{1 - n}$

For the area to be a constant,
$1 - n = 0 \Rightarrow n = 1$

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