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Thread: Astroid

  1. #1
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    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 $\displaystyle dy/dx=-(y/x)^1/3.$ 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.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by Hweengee View Post
    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 $\displaystyle dy/dx=-(y/x)^1/3.$ 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. Let $\displaystyle M(x_0,y_0)$ be a point of the astroid. The equation of the tangent line at $\displaystyle M$ is

    $\displaystyle y=-\left( \frac{y_0}{x_0}\right)^\frac13(x-x_0)+y_0$
    But how do i show that the length of this line passing through (q,0) and (0,p) is constant?
    Let $\displaystyle P(0,p)$ and $\displaystyle Q(0,q)$ be the intersection point of the tangent line with the y-axis and the x-axis, respectively. $\displaystyle p$ satisfies

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

    and $\displaystyle q$ is such that

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

    You can now solve these two equations for $\displaystyle p$ and $\displaystyle q$ and compute $\displaystyle PQ=\sqrt{p^2+q^2}$ to check that this length is constant. (remember that we have $\displaystyle x_0^\frac23+y_0^\frac23=a^\frac23$)
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  3. #3
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    Hello, Hweengee!

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

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


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


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

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

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


    The length of $\displaystyle AB$ is given by:

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


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

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

    . . . . . $\displaystyle = \;\bigg[-q^{\frac{1}{3}}\left(p^{\frac{2}{3}} + q^{\frac{2}{3}}\right)\bigg]^2\,\left(\frac{p^{\frac{2}{3}} + q^{\frac{2}{3}}}{q^{\frac{2}{3}}}\right) \;=$ .$\displaystyle 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: . $\displaystyle \overline{AB}^2 \;=\;\left(p^{\frac{2}{3}} + q^{\frac{2}{3}}\right)^3 \qquad\hdots \text{ but }p^{\frac{2}{3}} + q^{\frac{2}{3}} \:=\:a^{\frac{2}{3}}$


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

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