The quadratix that I know of is used to dissect angles, and is defined in a manner which makes its use in this area obvious.

Draw a square. Call the bottom left hand corner O, the origin. Now imagine the upper side on the square sweeping down the square at a constant rate until it reaches the bottom side. At the same time imagine a line passing through O, starting on the left hand side of the square, and rotating at a constant rate around O until it reaches the bottom side of the square. Now imagine these two motions starting and finishing at the same time.

The two lines will always cross at some point within the square, and the locus of these points (the curve you get when you join the points up) is the quadratix.

You can see that it will start at the top left hand corner of the square, and proceed across and down until it reaches the bottom side, about 2/3 the way along.

Now, dividing an angle is easy. Draw the angle you want to divided at O, with the base line the bottom of the square, and the other angle line directed into the square. This line will cut the quadratix at some point, so draw a line through this point parallel to the base of the square, call this line A B. Now dissect the line O A into the number of portions you wish to dissect the angle into. Draw horizontal lines through the points of dissection. These lines will meet the quadratix at certain points. Draw lines through these points to O, and you have dissected your angle.

It's obvious why this works from the definition of the quadratix. If you're having trouble seeing why, first consider trisecting the angle p/2, i.e. the corner of the square. Your line A B is now the top of the square, so trisect a vertical side of the square. So you will then have two horizontal lines, 1/3 and 2/3 of the way down the square, so where these meet the quadratix, the angle subtended at O will be 1/3 and 2/3 of the total angle, because that is how you drew the quadratix.

Now about those equations for the quadratix you mentioned earlier. To derive these it is easiest to take the square to be of side p/2, hopefully you can see why the length doesn't really matter. The quadratix will be all those points where the horizontal line, distance q from the bottom, and the line through O at angle q to the bottom meet. You can see that the choice of side p/2 meant we were able to use the same q for the angle of one line and the height of the other, if the square had been of side "a" we would have needed to include a scale factor in the height.

To represent the meeting of two lines algebraically, we should introduce a coordinate system. Call the base of the square the x-axis and the LHS the y-axis as is usual. Now the equation of the horizontal line at height q is y=q, and the equation of the line through O at angle qis y=x×tan(q).

For these to meet q = x×tan(q) and so x=qcot(q).

So we have a parametric eqn for the quadratix:

y=q x=q×cot(q) with 0 < q £ p/2

For a Cartesian equation we eliminate q:

x=q×cot(q)

Which is the equation you gave in your question, but with y and x transposed (just a reflection in y=x).

For a polar equation, we should go back to the diagram. Draw a square, a horizontal line and a line through O such that they meet on the quadratix, i.e. one has height q and the other has angle q. Now use trig. to find the distance from O to the point on the quadratix where they meet, call this distance r. We get sin(q)=q/r and hence: r=q×cosec(q). Which is the equation you gave.

Note that this method does not solve the famous "Trisection of the angle" problem, as the problem stipulates that it must be done with straight edge and compasses alone, and you can not construct a quadratix with these. It has been proven in fact that is is impossible to trisect an angle with just straight edge and compasses, so don't even bother trying.

Hope you followed all this,

Please write again,