Here is my task: An arbitrary ray $OE$ crosses the circle at point $D$:
$$x ^ 2 + (y – a / 2) ^ 2 = (a / 2) ^ 2$$
and a tangent to it(at point $E$), passing through point $C$, diametrically opposite to $O$.
Through points $D$ and $E$, straight lines are drawn parallel to the axes $Ox$ and $Oy$, respectively, up to the intersection at point M.
Equate the line formed by the $M$ points and draw it (Agnesi's curl).
My attempt
I think that this problem need using Polar coordinates. But I have no idea how to start. Maybe if we assume that $M$ has coordinates $(x,y)$ we can find this point(using polar coordinates like $x=r\sin(q)$ or something like that) but how to equate the line? And of course I have already drawn it. After an hour i have idea if we find point $M$ can we use $M$-coordinates to get parametric equation of a line? If yes, how to make first step? Ok, after a few time i found $x$ like a coordinate of a point $M$ (i use as a parameter angle between $OE$ and $OC$ (q) so $x=tgqa/cosq$) how can find $OB$?
Best Answer
Spoiler alert: this is not a hint, here is a solution using cartesian coordinates.
The calculation is based on similarity of triangles $OBD$ and $OCE$. First we consider the right side of the curve ($x \ge 0$) and then we note that the curve is symmetric. Our objective is to find $y_M$ as a function of $x_M$.
From the similarity of triangles $OBD$ and $OCE$ we have: $$\frac{OB}{OC} = \frac{BD}{CE} $$ $$\frac{y_M}{a} = \frac{x_D}{x_M} $$ $$\therefore y_M = a \frac{x_D}{x_M} \qquad(1)$$ Now, to find $x_D$ we note that $D$ is on the given circle and that $y_D = y_M$ : $${x_D}^2 + (y_M - \frac a2)^2 = (\frac a2)^2 $$ $$\therefore {x_D}^2 = y_M (a - y_M) \qquad(2)$$ By squaring equation (1) and using equation (2) we have: $$y_M = a^2 \frac{a-y_M}{{x_M}^2} $$ which can be rewritten as $$y_M = \frac{a^3}{a^2 + {x_M}^2} \qquad(3)$$ which is the equation of the loci of $M$ . Note that the curve of desired loci is symmetric, and equation (3) accounts for positive as well as negative values of $x$ .