general second degree equation(2 )

ax²＋2hxy＋by²＋2gx＋2fy＋c＝0　　　　　　　　　(1)

Let (p,q) be the required the intersection point.

Case 1 (a≠0, b≠0):

Putting y＝q into (1), we have

　　　ax²＋2hxq＋bq²＋2gx＋2fq＋c＝0
＝＞　ax²＋2(hq＋g)x＋(bq²＋2fq＋c)＝0　　　　　　(2)

Since there is only one intersection point, equation (2) has two coincident real roots (i.e. x＝p). Therefore,

Discriminant＝[2(hq＋g)]²－4a(bq²＋2fq＋c)＝0
＝＞　　(ab－h²)q²＋2(af－gh)q＋ac－g²＝0　　　　　(3)

and p is given by

p＝[－2(hq＋2g)±√0]/(2a)　　＝＞　　ap＋hq＝2g　　(4)

Putting x＝p into (1) and repeat the above argument, we have

hp＋bq＝2f　　　　　　　　　　　　　　　　　　　(5)

Solving equations (4) and (5) for p and q, we have

the intersection point (p,q)＝（(fh－bg)/(ab－h²),(gh－af)/(ab－h²)） 　(6)

and the condition ab－h²≠0　　　　　　　　　 　　(7)

Putting q＝(gh－af)/(ab－h²) into (3) to get another condition:

　　　　(gh－af)²/(ab－h²)－2(gh－af)²/(ab－h²)＋ac－g²＝0
＝＞　　－(gh－af)²＋(ac－g²)(ab－h²)＝0
＝＞　　－g²h²＋2afgh－a²f²＋a²bc－abg²－ach²＋g²h²＝0
＝＞　　abc＋2fgh－af²－bg²－ch²＝0　　　 　 (8)

Case 2 (a＝0, b≠0):

Setting a＝0, equation (1), (6), (7) and (8) become

2hxy＋by²＋2gx＋2fy＋c＝0　　　　　　　　　(9)
(p,q)＝（(bg－fh)/h²,－g/h）　　　　　　　　　(10)
h≠0　　　　　　　　　　　　　　　　　　　(11)
2fgh－bg²－ch²＝0　　　　　　　　　　　　　(12)

Putting (10) into (9), we have

－2g(bg－fh)/h²＋bg²/h²＋2g(bg－fh)/h²－2fg/h＋c
＝bg²/h²－2fg/h＋c＝－(2fgh－bg²－ch²)/h²＝0　　by (12)

This verifies the results (6), (7) and (8) from case 1 is also true for a＝0 and b≠0

Case 3 (a≠0, b＝0) and case 4 (a＝b＝0):

By the same method for case 2, the results (6), (7) and (8) can also be verified.

Summing up:
The intersection point (p,q)＝（(fh－bg)/(ab－h²),(gh－af)/(ab－h²)）
The conditions are: ab≠h² and abc＋2fgh－af²－bg²－ch²＝0

too bad

You are going through the harder way,
the easy way is like this:

equation for staight line : a x + b y + c = 0
As you can see, order for x and y are both 1,
multiplying them makes the second degree equation you provided.
But this Standard Form is not a good form actually,
y = m x + c is more convenient, as b is reduced( not vanish, just fit into the ratio)

For not being mixed with your parameters, i let the two equation be
r x + s y + t = 0
u x + v y + w = 0
Here, there is a conceptual question, what does it mean if r/s = u/v?
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..
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if that ratio is the same, then the slope is the same, which means they are parallel lines.
Parallel lines never intersect with each other!
you see? This is the assumption required.

r x + s y + t = 0 and u x + v y + w = 0 gives
( r x + s y + t ) ( u x + v y + w ) = 0
Expand it and compare coefficients, e.g. a= r*u, c= t*w
6 parameters to 6 parameters, so there must be a one-one correspondence.
To find the intersection point, solve the simultaneous equation:
{ r x + s y + t = 0 and u x + v y + w = 0 }

These last two steps are merely clumsy calculatoin, and i am leaving them to yourself =P
Lastly, i would like to know the source of this question? is it homework for some maths course?

2008-06-30 20:22:00 補充：
excuse me, are you refering to my answer?
can you give me more advise, myisland8132?

2008-07-01 20:35:15 補充：
nice work, nychan63 !

2008-07-01 20:37:51 補充：
i think the method i suggest is not as good, please just take it as a reference~