So I've tried to use the envelope formula to figure it out.
Using the intercept equation for a straight line graph:
$$\frac{x}{a}+\frac{y}{b}=1$$ and $$a^2+b^2=S^2, b=(S^2-a^2)^\frac{1}{2}$$
$$\frac{x}{a}+\frac{y}{(S^2-a^2)^\frac{1}{2}}=1$$
So now $a$ will be the variable. Using the forumla for envelopes, I have,
$$\frac{-x}{a^2}+\frac{ya}{(S^2-a^2)^\frac{3}{2}}=0$$
The next step is obviously to express $a$ into terms of $x$ and $y$, but when I try to do so, I get a polynomial with degree 6 (which is unsolvable). Am I missing out something here?
EDIT: As per an answer, I've expressed it in
$$y=\sqrt{S^2-a^2}-\frac{x\sqrt{S^2-a^2}}{a}$$
Differentiating both sides w.r.t $a$, I've obtained,
$$0=\frac{-a}{\sqrt{S^2-a^2}}+\frac{x}{\sqrt{S^2-a^2}}+\frac{x\sqrt{S^2-a^2}}{a^2}$$
Simplifying:
$$a=S^\frac{2}{3} x^\frac{1}{3}$$
However, I am unable to simply the expression after subbing in. The answer should be $$x^\frac{2}{3}+y^\frac{2}{3}=S^\frac{2}{3}$$
Write $$\frac{x}{a}+\frac{y}{(S^2-a^2)^\frac{1}{2}}=1$$ as $$y=\dfrac{a-x}{a}\sqrt{S^2-a^2}$$ then differentiate respect to $a$ gives $$0=(0+\dfrac{x}{a^2})\sqrt{S^2-a^2}+\dfrac{a-x}{a}\dfrac{-2a}{2\sqrt{S^2-a^2}}$$ or $xS^2=a^3$, then with $a=\sqrt[3]{xS^2}$ we find $$y=\left(1-\sqrt[3]{\dfrac{x^2}{S^2}}\right)\sqrt{S^2-S\sqrt[3]{x^2S}}$$