I have following two equations:
a*x^2 + b*x + c*y^2 + d*y + f= 0
g*x^2 + h*x + i*y^2 + j*y + k= 0
Can anyone solve for x and y? I am seeking for an algebraic/symbolic solution.
Thanks
Your system is equivalent to the following 2 equations. Note that the first one contains only y, and the second is linear in x. I found them by computing the Groebner basis.
(a^2*i^2-2*a*c*g*i+c^2*g^2)*y^4+(2*a^2*i*j-2*a*c*g*j-2*a*d*g*i+2*c*d*g^2)*y^3+(2*a^2*i*k+a^2*j^2-a*b*h*i-2*a*c*g*k+a*c*h^2-2*a*d*g*j-2*a*f*g*i+b^2*g*i-b*c*g*h+2*c*f*g^2+d^2*g^2)*y^2+(2*a^2*j*k-a*b*h*j-2*a*d*g*k+a*d*h^2-2*a*f*g*j+b^2*g*j-b*d*g*h+2*d*f*g^2)*y+a^2*k^2-a*b*h*k-2*a*f*g*k+a*f*h^2+b^2*g*k-b*f*g*h+f^2*g^2 = 0
(a*h-b*g)*x+(a*i-c*g)*y^2+(a*j-d*g)*y+a*k-f*g = 0
Your system is equivalent to the following 2 equations. Note that the first one contains only y, and the second is linear in x. I found them by computing the Groebner basis.
(a^2*i^2-2*a*c*g*i+c^2*g^2)*y^4+(2*a^2*i*j-2*a*c*g*j-2*a*d*g*i+2*c*d*g^2)*y^3+(2*a^2*i*k+a^2*j^2-a*b*h*i-2*a*c*g*k+a*c*h^2-2*a*d*g*j-2*a*f*g*i+b^2*g*i-b*c*g*h+2*c*f*g^2+d^2*g^2)*y^2+(2*a^2*j*k-a*b*h*j-2*a*d*g*k+a*d*h^2-2*a*f*g*j+b^2*g*j-b*d*g*h+2*d*f*g^2)*y+a^2*k^2-a*b*h*k-2*a*f*g*k+a*f*h^2+b^2*g*k-b*f*g*h+f^2*g^2 = 0
(a*h-b*g)*x+(a*i-c*g)*y^2+(a*j-d*g)*y+a*k-f*g = 0
In the plane, if a or c, g or i are non-zero, each of two equations defines one of following objects: circle, ellipse, hyperbole, one or two lines, or null set. It depends on all coefficients in the equation.
Could you tell us the values of a,b,...,k ? The solution (from 0 to 4 ordered pairs (x,y), or infinity) can be not only computed but also nicely visualized.
Could you tell us what do you mean by "solving" ? What kind of information do you expect on the solutions since this information clearly depends on some indeterminate that might be considered as parameters.
in these equations, a,b,...,k all are constants while x and y are variables. i wanted to know the value of x (or y) in terms of these constants.
You will get quite enormous expressions ... either by hand or automatically.
Automatically, Maple will give you the answer using the command allvalues(solve({q1,q2},{x,y})).
but it is ... enormous.
If you want to check why it is so enormous, do almost the same computation as Maple by hand : compute the resultant of the two equations wrt x and then you vill get a polynomial of degree 4 in y. You can then express all its roots (which are then all the y coordinates when a and g are not null) wrt the parameters using radicals supposing that the leading coefficient is not null. It is then sufficient to plug each of them in the two initial equations in x and again solve them manually (they both are of degree 2).
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