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I've got the translation sorted but I'm all tied up with the rotation at the moment.
It sounds like you have both steps, in which case cannot you just convert the algorithm into a formula for the foci?
Thanks, gnitsuk and others. I just registered in this forum because I was looking for a solution for exactly this problem.
To futher complete this topic I'll post that I now have a solution for the hyperbola.
Find the foci.
(If the equation represents a parabola then one should find the focus and directorix, but case is solved above).
If the equation represents a hyperbola, which will be the case when
then the method is as follows:
Find the asymptotes of the hyperbola
case B != 0.0:
Gradients given by
case B = 0.0:
Gradients given by and other asymptote vertical
Intercetps by and
This covers all cases of the hyperbola.
Now, find intersection points of these two lines, this is the centre of the hyperbola.
Next find the two lines which bisect the two angles formed by the two asymptotes. Only one of these lines will intersect the hyperbola, the two intersection points are the vertices of the hyperbola.
Let a = the distance from the centre to a vertex.
Next, move from the centre a distance ae along the line joining the two vertices towards each vertex where e is the eccentricity of the hyperbola calculated as:
where n is 1 or -1 as the determinant:
is negative or positive respectively.
This gives the two foci of the hyperbola.
I've got this running nicely in an interactive application and it works very well.
If anyone can think of a more elegant method I'd be very glad to hear.
Onward to the ellipse..........
Thanks for your suggestion.
How can one determine the correct change of reference frame that will make the coordinate axes become the lines of symmetry of the hyperbola?
Thanks for any help.
Hello. I wonder could anyone help me with the following.
Let's start with the parabola. This is the curve comprised of the locus of points whose distance to a line ax+by+c=0 (the directorix) is always the same as their distance to a point (u,v) (the focus).
So, given the directorix and the focus can one derive the general form given above for that particular parabola?
Yes. One can equate the distance of the general point (x,y) from the directorix and the focus like so:
Now one just moves everything to the left hand side, collects up terms in x^2,y^2,xy,x and y and obtains the results:
Good. So given u,v,a,b,c as input, one can calculate A,B,C,D,E and F as output.
How about the inverse. Given a general conic (A,B,C,D,E,F) known the be a parabola, can one find the focus and directorix.
Yes. From the first two equations above, we have:
Now this alone is not sufficient to determine a and b, as each square root may be either positive or negative, but we also have the third equation above. So if we calculate the positive roots and then find that -2ab is not equal to C then we can choose either a or b and reverse its sign and we are done. So we now know a and b. The remaining equations give as c,u and v as:
My question is, how about the hyperbola and the ellipse. Can we do the same thing?
Let's just consider the hyperbola.
The first part is easy enough. Given the foci ( two of them obviously, (a,b) and (c,d) ) of the hyperbola we can set the difference of the distances from the general point (x,y) on the hyperbola to these foci to be a constant (say k) to obtain:
Now again, we collect powers of x and y etc. and after the dust has settled we obtain:
This all works perfectly well. I've written code to allow me to move the mouse around on the screen to define the position of the second focus given the first and the hyperbola is drawn from the calculated values of A,B,C,D,E and F and all is well.
BUT, how about obtaining a,c,b,d from A,B,C,D,E and F. That looks hard to me.
As you can see, I've been pleasantly busy. So after all this my question is:
Given the general equation of a hyperbola or ellipse. How to find the foci?
I have solved the problem fully for the parabola as you see above. But only half the problem for the hyperbola and ellipse (i.e. given the foci I can obtain the general equation).
Thanks for any suggestions. There may be a much simpler approach which I am missing.