# Astroid

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• September 14th 2008, 02:47 AM
Hweengee
Astroid
Got kind of stuck on this question.

Suppose a is a non-zero constant. x^2/3+y^2/3=a^2/3. Show that the length of the portion of any tangent line to this astroid cut off by the x and y axes is constant.

I've differentiated implicitly to obtain the gradient of the tangent as $dy/dx=-(y/x)^1/3.$ And have obtained the equation of the tangent line passing through (q,0) and (0,p) as y=-(y/x)^1/3.x+p.

But how do i show that the length of this line passing through (q,0) and (0,p) is constant?

Thanks in advance.
• September 14th 2008, 08:04 AM
flyingsquirrel
Hello,
Quote:

Originally Posted by Hweengee
Got kind of stuck on this question.

Suppose a is a non-zero constant. x^2/3+y^2/3=a^2/3. Show that the length of the portion of any tangent line to this astroid cut off by the x and y axes is constant.

I've differentiated implicitly to obtain the gradient of the tangent as $dy/dx=-(y/x)^1/3.$ And have obtained the equation of the tangent line passing through (q,0) and (0,p) as y=-(y/x)^1/3.x+p.

There are many x's and y's in the equation of the line. (Wondering) Let $M(x_0,y_0)$ be a point of the astroid. The equation of the tangent line at $M$ is

$y=-\left( \frac{y_0}{x_0}\right)^\frac13(x-x_0)+y_0$
Quote:

But how do i show that the length of this line passing through (q,0) and (0,p) is constant?
Let $P(0,p)$ and $Q(0,q)$ be the intersection point of the tangent line with the y-axis and the x-axis, respectively. $p$ satisfies

$p=-\left( \frac{y_0}{x_0}\right)^\frac13(0-x_0)+y_0$

and $q$ is such that

$0=-\left( \frac{y_0}{x_0}\right)^\frac13(q-x_0)+y_0$

You can now solve these two equations for $p$ and $q$ and compute $PQ=\sqrt{p^2+q^2}$ to check that this length is constant. (remember that we have $x_0^\frac23+y_0^\frac23=a^\frac23$)
• September 14th 2008, 11:51 AM
Soroban
Hello, Hweengee!

Quote:

Suppose $a$ is a non-zero constant, and: . $x^{\frac{2}{3}} + y^{\frac{2}{3}}\:=\:a^{\frac{2}{3}}$

Show that the length of the portion of any tangent line to this astroid
cut off by the x- and y- axes is constant.

Your derivative is correct: . $\frac{dy}{dx} \:=\:-\left(\frac{y}{x}\right)^{\frac{1}{3}} \:=\:-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$

Let $P(p,q)$ be any point on the astroid.
The slope of the tangent at $P$ is: . $m \;=\;-\frac{q^{\frac{1}{3}}} {p^{\frac{1}{3}}}$

The tangent through $P(p,q)$ with slope $m$
. . has the equation: . $y - q \;=\;m(x - p) \quad\Rightarrow\quad y \;=\;mx + q - mp$

Its x-intercept is: . $A\left(\frac{mp-q}{m},\;0\right)$

Its y-intercept is: . $B\bigg(0,\;-(mp-q)\bigg)$

The length of $AB$ is given by:

. . $\overline{AB}^2 \;=\;\left(\frac{mp-q}{m}\right)^2 + (mp-q)^2 \;=\;(mp-q)^2\left(\frac{1}{m^2} + 1\right)$

Replace $m$ with $-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}$

. . $AB^2 \;=\;\left(-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}\!\cdot\! p \:- \:q\right)^2\,\left(\frac{p^{\frac{2}{3}}}{q^{\fra c{2}{3}}} +1\right) \;=$ . $\left(-p^{\frac{2}{3}}q^{\frac{1}{3}} - q\right)^2\,\left(\frac{p^{\frac{2}{3}} + q^{\frac{2}{3}}}{q^{\frac{2}{3}}}\right)$

. . . . . $= \;\bigg[-q^{\frac{1}{3}}\left(p^{\frac{2}{3}} + q^{\frac{2}{3}}\right)\bigg]^2\,\left(\frac{p^{\frac{2}{3}} + q^{\frac{2}{3}}}{q^{\frac{2}{3}}}\right) \;=$ . $q^{\frac{2}{3}}\left(p^{\frac{2}{3}} + q^{\frac{2}{3}}\right)^2\cdot\frac{p^{\frac{2}{3}} + q^{\frac{2}{3}}}{q^{\frac{2}{3}}}$

Hence: . $\overline{AB}^2 \;=\;\left(p^{\frac{2}{3}} + q^{\frac{2}{3}}\right)^3 \qquad\hdots \text{ but }p^{\frac{2}{3}} + q^{\frac{2}{3}} \:=\:a^{\frac{2}{3}}$

Therefore: . $\overline{AB}^2 \:=\:\left(a^{\frac{2}{3}}\right)^3 \:=\:a^2 \quad\Rightarrow\quad \overline{AB} \:=\:a \quad\hdots \text{ Q.E.D.}$