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How can I prove that an astroid is an envelope of all line segments of length 1 from the x-axis to the y-axis?

I read one proof of this online at the link but I don't understand how this proof works.

Therefore, it would be much appreciated if someone could show me another proof or make the online proof more understandable. For example, for the first step, why is it x*/t^2*+y/(a^2-t^2)? I mean, I get that this is because of the Pythagorean theorem, but why are these the denominators of x and y?

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I was having trouble with understanding the first step; I am not sure why the first step is x/t^2+y/(a^2-t^2) and why F(x,y,t) is that equation. Could you please explain the first step to me? Thanks. Also, what does wrt mean? – Jake Perentosh May 29 '13 at 4:37
Maybe review – Amzoti May 29 '13 at 5:03
ok it helps a bit but I get stuck in the middle... can we just go back to the original proof on Why is F(x,y,t) = x/t^2+y/(a^2-t^2)^0.5? – Jake Perentosh May 29 '13 at 5:36
There is a square root term in the second expression, are you missing it or I am looking at the wrong thing? – Amzoti May 29 '13 at 5:37
Sorry I meant to write the square root... and I edited it. So why does F(x,y,t) = x/t^2+y/(a^2-t^2)^0.5 = 1? – Jake Perentosh May 29 '13 at 5:41

2 Answers 2

For a reverse proof, compute the tangent of the astroid, find the intersection with the axis and show they distance is 1:

Start with the parametric equation $(x,y)=(\cos^3\theta, \sin^3\theta)$, whose derivative relative to $\theta$ is $(x',y')=(3 \cos^2\theta \sin\theta, -3 \sin^3\theta \cos\theta)$. The tangent at $\theta$ has equation $\frac {x - \cos^3\theta} {3 \cos^2\theta \sin\theta} = \frac {x - \sin^3\theta} {-3 \cos\theta \sin^2\theta}$, which simplifies in $x \cos\theta + y \sin\theta= \cos\theta \sin\theta$. The tangent cuts the axis in $P=(0, \cos\theta)$ and $Q=(\sin\theta, 0)$. As $PQ=1$, this proves that the tangent of the astroid at any point is a segment of length 1 from the x-axis to the y-axis.

For the constructive proof, take the segment computes its equation $y=F_\theta(x)$, and, accordingly to the books, solve the system $y=\frac {\partial F_\theta} {\partial \theta} (x)$, which express that you choose the point on the segment in such y way that the segment is tangent to the envelope.

Take $P=(0, \cos\theta)$ and $Q=(\sin\theta, 0)$, the segment equation is $\frac x {\cos\theta} + \frac y {\sin\theta}=1$, which symplifies in $x \cos\theta + y \sin\theta= \cos\theta \sin\theta$ (as above). Derivating (relative to $\theta$!) leads to $-x \sin\theta + y \cos\theta= \cos^2\theta - \sin^2\theta$. So, the equation of the envelope is the solution of the system:

$\begin{cases} +x \cos\theta + y \sin\theta &= \cos\theta \sin\theta\\ -x \sin\theta + y \cos\theta &= \cos^2\theta - \sin^2\theta \end{cases}$

To solve the system multiply the first line by $\sin\theta$, the second by $\cos\theta$ and add to eliminate $x$, which leaves $y=\sin^3\theta$. Then multiply the first line by $\cos\theta$, the second by $\sin\theta$ and add to eliminate $y$, which leaves $x=\cos^3\theta$. So the envelope parametric equation is $(x,y)=(\cos^3\theta, \sin^3\theta)$. You can eliminate the parameter by replacing $(\cos\theta, \sin\theta)=(x^\frac 1 3, y^\frac 1 3)$ in $\cos^2 \theta+\sin^2\theta=1$ to get the familiar $x^\frac 2 3 + y^\frac 2 3 =1$.

I may seem quick with mental calculations, but a lot of simplifications will appear a long the way, as long as you keep your formulas as symmetric as possible.

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Here, I give a complete proof. Let $y=f(x)$ be the curve constructed by all line segments joining points in $x$-axis and points in $y$-axis of length $1$. Then, the value $f(a)$ is given by $$ f(a)=\sup_{0<p<1}\sqrt{1-p^2}(1-a/p) $$

By elementary calculus(differenting the right hand side of above with respect to $p$) one can see that it attains maximum at $p=\sqrt[3]{a}$. Putting $p=\sqrt[3]{a}$ above we have $$ f(a)=(1-a^{2/3})^{3/2}. $$

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