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Math Help - Descartes and mean proportionals

  1. #1
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    Unhappy Descartes and mean proportionals

    Descartes pointed out that the curves x^2=ay, x^2+y^2=ay+bx intersect in a point (x,y) such that x and y are the two mean proportionals between a and b. Show this.
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  2. #2
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    Hello, sjenkins!

    Descartes pointed out that the curves: . \begin{array}{cccc}x^2 &=& ay & {\color{blue}[1]} \\ x^2+y^2 &=& ay+bx & {\color{blue}[2]}\end{array}

    intersect in a point (x,y) such that x and y are the two mean proportionals

    between a and b. .Show this.
    We must show that . a, x, y, b form a geometric progression.


    From [1], we have: . x^2 \:=\:ay \quad\Rightarrow\quad x \:=\:\sqrt{ay}\;\;{\color{blue}[3]}


    Substitute into [2]: . ay + y^2 \:=\:ay + b\sqrt{ay} \quad\Rightarrow\quad y^2 \:=\:b\sqrt{ay}


    Square both sides: . y^4 \:=\:b^2\!\cdot\!ay \quad\Rightarrow\quad y^4 - ab^2y \:=\:0


    . . y\left(y^3-ab^2\right) \:=\:0\quad\Rightarrow\quad \begin{Bmatrix}{\color{red}\rlap{//////}}y \:=\: 0 \\ y^3 \:=\: ab^2\;\;\Rightarrow\;\;y \:=\: a^{\frac{1}{3}}b^{\frac{2}{3}} \end{Bmatrix}

    Disregarding the trivial solution, we have: . \boxed{y \:=\:a^{\frac{1}{3}}b^{\frac{2}{3}}}

    Substitute into [3]: . x \;=\;\sqrt{a\!\cdot\!a^{\frac{1}{3}}b^{\frac{2}{3}  }} \;=\;\sqrt{a^{\frac{4}{3}}b^{\frac{2}{3}}} \quad\Rightarrow\quad\boxed{ x \:=\:a^{\frac{2}{3}}b^{\frac{1}{3}}}


    The progression is: . \boxed{ {\color{blue}a,\;\;\sqrt[3]{a^2b},\;\;\sqrt[3]{ab^2},\;\;b}}

    . . The common ratio is: . r \:=\:\sqrt[3]{\frac{b}{a}}

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