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How do I find the center of a central conic with the equation [tex]ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0[/tex] Is there some easy forumlua for both the ellipse and hyperbola? Thanks!
If you succeed in rotating, you will not have any terms that involve xy products. If you had something of the formmattmns said:Also, what is translation?
Thanks
I think there is a better way. It is easier to visualize for the case of an ellipse. Think about what it means for dy/dx = 0 and dy/dx approaches infinity for an ellipse, and how you might use the location of a pair of points that satisfy either one of those conditions to locate the center of the ellipse. How could you find the points (x,y) that satisfy those conditions? Then relate those points to the zeros of the partial derivatives.mattmns said:Thanks for the clarification.
I am beginning to think the approach I was trying is not a good one.
Here is what I am trying to prove.
[tex]\phi (x,y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0[/tex]
for [tex] ab - h^2 \neg 0[/tex] That is: ab - h^2 not equal to 0
Show that the center of the central conic [tex]\phi (x,y) [/tex] is the interesection of the lines [tex] \frac{\partial \phi}{\partial x} = 0 [/tex] and [tex] \frac{\partial \phi}{\partial y} = 0 [/tex]
Do you think the approach I am trying, by first finding the center of the central conic is a good route? Thanks
The correspondence between the dy/dx results and the partial derivatives is exactly what you want. Think of any ellipse you can imagine, centered anywhere, with any aspect ration, with axes at any angle of rotation relative to the x-y axes. The solution to [tex]dy/dx = 0[/tex] is a line that will intercept that ellipse at two special points. The solution to [tex]dy/dx = \infty [/tex] is a line that intercepts the ellipse at two other special points. Think about what these points mean, and where these lines must intersect based on the symmetry of an ellipse.mattmns said:That sounds like a great idea, and I tried it, but I did not come up with exactly what I wanted.
I got dy/dx, using implicit differentiation, to be: [tex]\frac{-2ax - 2hy - 2g}{2hx + 2by + 2f}[/tex] So, when dy/dx = 0, then -2ax - 2hy - 2g = 0. You could also say from that, that when dy/dx = 0, then 2ax + 2hy + 2g = 0. Also, when dy/dx = [tex] \infty [/tex] then 2hx + 2by + 2f = 0.
Now, when I did the partial differentiation, I got the same answers, which is why I did not like it.
I got
[tex] \frac{\partial \phi}{\partial x} = 2ax + 2hy + 2g [/tex]
and
[tex] \frac {\partial \phi}{\partial y} = 2hx + 2by + 2f [/tex]
From here I am stuck, I am unsure of really what to do. It seems like it is close, but just not there.
I guess you could say that for the first part, dy/dx = 0, that [tex]y = \frac{-ax - g}{h}[/tex] Which is a line, but from there I am uncertain, other than finding the other line, which is [tex] y = \frac {-hx - f}{b}[/tex]