Tangential Network 1

The derivative of the equation $x^n + y^n = 1$ is

$n\left(x^{n-1}\:dx+y^{n-1}\:dy\right) = 0.$

We use the general equation for a line in the form $Ax+By=C$ , where the slope is $-A/B$ . For a tangent line this slope must be equal to $-A/B = dy/dx$ , which we may rewrite as $A\:dx + B\:dy = 0$ . So the tangent line to point $(a,b)$ has the equation

$a^{n-1}x + b^{n-1}y = C,$

and plugging in $x = a, y = b$ shows that in fact, $C = 1$ .

The intersection points with the axes are $(\xi,0)$ and $(0,\eta)$ , with $\xi = \frac C{a^{n-1}} = a^{1-n};\ \ \eta = \frac C{b^{n-1}} = b^{1-n}.$

We know that $a^n + b^n = 1$ , and we want to have $\xi + \eta = a^{1-n} + b^{1-n} =$ constant. Clearly, this will be the case if $n = 1-n$ . It follows that $\boxed{n = \tfrac12}$ .

(Another way of viewing this problem: if $\xi + \eta = 1$ , then the lines connecting $(\xi,0)$ and $(0,\eta)$ sweep out a curve with the equation $\sqrt x + \sqrt y = 1$ . This curve looks a bit like a half-circle, but isn't really one...)