Locus of points - parabola (2)

[Loki123]

Hmmm, having constructed my diagram (above @ Post #15) and given some further thought to the matter, in light of that, I think I can now see how the answer given (y²=½px), actually does work; though I have absolutely no idea of how to arrive at it with just the information provided in the original question (as first posted).

I constructed the table below (qv) which, may not exactly prove the result but certainly appears to verify it. Please see the attached table in which I have considered the range of y-values that produce the perfect squares only.

It would seem, therefore, that for any given parabola of the form y²=½px the points (x, p) & (x,-p) will, indeed, lie on the ordinate (vertical line) x=2p so p does give the magnitude of the y-values (ordinates?) sought.

Does that make any sense???

(I still suspect that something has been "lost in translation" however: "in the middle(?) of the ordinate" still bamboozles me!)

I found another question with that:

Determine the geometric locus of the points that divide the ordinates of the points of the parabola y ^ 2 = 2px in the scale m: n, where the first segment is closer to the abscissa axis, and the second farther away from it.

Solution is:

Y^2=(2pm^2X)/(m+n^2)

This question and the question this post is initially about both have x, y as capital letters in the solution, they are the only ones to have this. They are the only ones to use the ordinates of points too.

 

[Dr.Peterson]

I found another question with that:

Determine the geometric locus of the points that divide the ordinates of the points of the parabola y ^ 2 = 2px in the scale m: n, where the first segment is closer to the abscissa axis, and the second farther away from it.

Solution is:

Y^2=(2pm^2X)/(m+n^2)

This question and the question this post is initially about both have x, y as capital letters in the solution, they are the only ones to have this. They are the only ones to use the ordinates of points too.

Here, it appears that "the ordinate of a point" means the vertical segment from the x axis to the point, and dividing that segment in the ratio m:n essentially means multiplying y by m/(m+n). So if you meant y^2=(2pm^2x)/((m+n)^2), the answer makes sense, as it is equivalent to $y=\pm\frac{m}{m+n}\sqrt{2px}$.