The parabola $y^2 = 3x$ is given. Two perpendicular straight lines are drawn from the origin point, intersecting the parabola in points P and Q. Find equation (in cartesian form) of the set of centers of mass of all triangles OPQ (where O is the origin point).
Please help, i managed to notice, that one line is $ax$, the second is $-\frac{1}{a}x$. Also, the centers of mass move on the shape of another parabola as we rotate the lines.
On
Step 1: calculate points $P$ and $Q$. You already have the equation for those lines, make them intersect with the parabola. Assume $a>0$. Note that $x>0$ $$ax=\sqrt{3x}$$ yields $$x=\frac{3}{a^2}$$ and $$y=\frac{3}{a}$$ The second equation is $y=-x/a$. You square it and get $$x=3a^2$$ and $$y=-3a$$ Step 2: The center of mass is at the average position of the vertices $(0,0)$, $(3/a^2,3/a)$, $(3a^2,-3a)$
On
A generic line through the origin has equation $r:y=mx$ and we have $r^{\perp}:y=-\frac{1}{m}x$
The intersection of $r$ and $r^{\perp}$ with the parabola $y^2=3x$ are respectively $P\left(\frac{3}{m^2};\;\frac{3}{m}\right)$ and $Q(3m^2,\;-3m)$
The solution of the systems $ \left\{ \begin{array}{l} y^2=3x \\ y=mx \\ \end{array} \right.\quad\quad $ $ \left\{ \begin{array}{l} y^2=3x \\ y=-\frac{1}{m}\,x \\ \end{array} \right. $
Centroid $C$ of triangle $OPQ$ has coordinates which are the average of the coordinates of $(O,P,Q)$ that is $C\left(\frac{1}{m^2}+m^2,\frac{1}{m}-m\right)$
Point $C$ as $m$ varies describes a curve of parametric equations
$ \left\{ \begin{array}{l} x=m^2+\frac{1}{m^2} \\ y=\frac{1}{m}-m \\ \end{array} \right. $
To eliminate the parameter let's square $y$
$\left(\frac{1}{m}-m \right)^2=\frac{1}{m^2}+m^2-2$
so we get the equation
$y^2=x-2$
Hope this is useful $$...$$
let your first line is $y=mx$ thus second line is $y=-x/m$ now you can easily find intersection points P and Q
P($\frac{3}{m^2}$,$\frac{3}{m}$) similiarly Q($3m^2$,$-3m$)
now centroid for right angle triangle is point ($\frac{OP}{3}$,$\frac{OQ}{3}$)