# Thread: Descartes and mean proportionals

1. ## 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.

2. Hello, sjenkins!

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

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

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

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

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

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

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

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

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

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

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