Using == instead of = does not appear to make a difference in this case.
Here is another approach yet without success:
1 Isolate the 'y' coordinate from the quadratic Hyperbola equation in Wolfram:
Solve[{A1 x^2 + B1 x y + C1 y^2 + D1 x + S1 y + F1 == 0}, {y}]
2 Result from above:
y = (-sqrt((B1 x+S1)^2-4 C1 (A1 x^2+D1 x+F1))-B1 x-S1)/(2 C1) and C1!=0
3 I do the same for the other Hyperbola
y = (-sqrt((B2 x+S2)^2-4 C2 (A2 x^2+D2 x+F2))-B2 x-S2)/(2 C2) and C2!=0
4 I put the two together:
(-sqrt((B1 x+S1)^2-4 C1 (A1 x^2+D1 x+F1))-B1 x-S1)/(2 C1) == (-sqrt((B2 x+S2)^2-4 C2 (A2 x^2+D2 x+F2))-B2 x-S2)/(2 C2)
5 I solve for x in Wolfram:
Solve [{(-sqrt((B1 x+S1)^2-4 C1 (A1 x^2+D1 x+F1))-B1 x-S1)/(2 C1) == (-sqrt((B2 x+S2)^2-4 C2 (A2 x^2+D2 x+F2))-B2 x-S2)/(2 C2)}, {x}]
6 Result from above:
Wolfram|Alpha doesn't understand your query.
Showing instead result for query: Solve sqrt x^2