If the tangent at any point on the curve $x^4+y^4=c^4$ cuts off intercepts a and b on the coordinate axes, the value of $a^{-4 / 3}+b^{-4 / 3}$ is

$c^{-4 / 3}$

Let $\left(x_1, y_1\right)$ be a point on the given curve.

We have,

$x^4+y^4=c^4 \Rightarrow 4 x^3+4 y^3 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x^3}{y^3}$

$\Rightarrow \left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}= \frac{-x_1^3}{y_1^3}$

The equation of the tangent at $\left(x_1, y_1\right)$ is

$y-y_1=\frac{-x_1{ }^3}{y_1{ }^3}\left(x-x_1\right)$

$\Rightarrow y y_1{ }^3-y_1{ }^4=-x x_1{ }^3+x_1{ }^4$

$\Rightarrow x x_1{ }^3+y y_1{ }^3=x_1{ }^4+y_1{ }^4$

$\Rightarrow x x_1{ }^3+y y_1{ }^3=c^4$           $\left[∵  \left(x_1, y_1\right) \text { lies on } x^4+y^4=c^4\right]$

The lengths of intercepts made on the coordinate axes by this tangent are $c^4 / x_1{ }^3$ and $c^4 / y_1{ }^3$ respectively.

∴  $a=\frac{c^4}{x_1{ }^3}$ and $b=\frac{c^4}{x_1{ }^3}$

$\Rightarrow \quad x_1{ }^3=\frac{c^4}{a}$ and $y_1{ }^3=\frac{c^4}{b} \Rightarrow x_1=\left(\frac{c^4}{a}\right)^{1 / 3}$ and $y_1=\left(\frac{c^4}{b}\right)^{1 / 3}$

Since $\left(x_1, y_1\right)$ lies on $x^4+y^4=c^4$. Therefore, $x_1^4+y_1{ }^4=c^4$

$\Rightarrow \left(\frac{c^4}{a}\right)^{4 / 3}+\left(\frac{c^4}{b}\right)^{4 / 3}=c^4 \Rightarrow a^{-4 / 3}+b^{-4 / 3}=c^{-4 / 3}$