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mattmns
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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!
mattmns 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
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]
The center of a central conic is the point that lies at the intersection of the major and minor axes of the conic. It is also the point where all the diameters of the conic intersect.
The center of a central conic can be determined by finding the midpoint of the major and minor axes, or by using the equation of the conic and solving for the x and y coordinates of the center.
The center of a central conic is significant because it is the point of symmetry for the conic. This means that any point on one side of the center has a corresponding point on the other side with the same distance from the center.
The center of a central conic determines the shape of the conic. If the center is located at the origin, the conic will be a circle. If the center is off-center, the conic will be an ellipse. And if the center is outside the conic, it will be a hyperbola.
Yes, the center of a central conic can have negative coordinates. This will result in a conic that is shifted to a different position on the coordinate plane, but the shape of the conic will remain the same.