Why post it 3 times...expect 3 answers ?
Seems to be limitless solutions.
Here's one of them:
a=1, b=1, c=2, p=1, q=2, r=2, x=1, y=1
Any comments?
I have two equations with two unknowns x and y.
(r-yp)x^2 –{y^2 (b-a)-2yq+(a+c)}x+(y^2+1)( yp -r)=0 … … (1)
(q-xp)y^2 –{x^2 (a-b)-2xr+(b+c)}y+(x^2+1)( xp -q)=0 … … (2)
here a,b,c,p,q,r are some constants.
How to solve this ? Pls help me.
All I can see is this:
Your equations:
(r-yp)x^2 – [y^2 (b-a) - 2yq + (a+c)]x + (y^2+1)(yp-r) = 0 … … (1)
(q-xp)y^2 – [x^2 (a-b) - 2xr + (b+c)]y + (x^2+1)(xp-q) = 0 … … (2)
I'm replacing p,q,r with d,e,f:
(f-dy)x^2 – [y^2 (b-a) - 2ey + a+c]x + (y^2+1)(dy-f) = 0 … … (1)
(e-dx)y^2 – [x^2 (a-b) - 2fx + b+c]y + (x^2+1)(dx-e) = 0 … … (2)
Let u = f-dy ; then -u = dy-f
Let v = e-dx ; then -v = dx-e
Let w = b-a ; then -w = a-b
ux^2 – (wy^2 - 2ey + a + c)x - u(y^2 + 1) = 0 … … (1)
vy^2 – (-wx^2 - 2fx + b + c)y - v(x^2 + 1) = 0 … … (2)
Rewriting (2):
(wy - v)x^2 + (2fy)x + v(y^2 - 1) - y(b + c) = 0 (3)
So we now have:
ux^2 – (wy^2 - 2ey + a + c)x - u(y^2 + 1) = 0 … … (1)
(wy - v)x^2 + (2fy)x + v(y^2 - 1) - y(b + c) = 0 ... ...(3)
Both can be solved for x using the quadratic formula.
I'll stop here for now.
Follow all that? Any comments?