Determine the number of solutions of nonlinear system without solving. $x^2-y^2+2y=0$,
$2x+y^2-6=0$
I need to determine the number of solutions without solving it. There is a hint that a graph can help but I am still not sure how to go about this.
Thanks
 A: If this question were being asked in a context where we could not make use of a graphing calculator or program, we might approach it in this way.  The equations represent curves that are both conic sections.  The first can be rearranged, after completing the square, into
$$ x^2 \ - \ y^2 \ + \ 2y \ = \ 0 \ \ \Rightarrow \ \ x^2 \ - \ ( \ y^2 \ - \ 2y \ + \ 1 \ ) \ = \ 0 \ - \ 1 \ \ \Rightarrow \ \ (y \ - \ 1)^2 \ - \ x^2 \ = \ 1 \ \ , $$
which is the equation of a "vertical" hyperbola with its center at $ \ ( 0, \ 1) \ $ .  The second equation can be written as
$$ 2x \ + \ y^2 \ - \ 6 \ = \ 0 \ \ \Rightarrow \ \ y^2 \ = \ -2x \ + \ 6 \ \ \Rightarrow \ \ y^2 \ = \ -2 \ (x \ - \ 3) \ \ , $$
which represents a "horizontal" parabola which "opens to the left" and has its vertex at $ \ (3, \ 0) \ $ .
We can consider this from a purely geometrical point of view.  The branches of the hyperbola are oriented so that they open "upward" and "downward" parallel to the $ \ y-$ axis (in fact, the $ \ y-$ axis is the symmetry axis); it is an "open" curve, so the branches extend infinitely.  Likewise, the parabola is "open", with its "arms" extending infinitely "to the left".
If the vertex of the parabola lies to the left of the center of the hyperbola, it is likely that it will never intersect the hyperbola, since the parabola's "arms" become shallower and shallower in slope as they stretch away from the vertex, while the branches of the hyperbola approach the constant slopes of its asymptotes.  On the other hand, in the situation we have with the vertex of the parabola to the right of the hyperbola's center, it is very likely that the arms of the parabola will cross both asymptotes of the hyperbola, twice for each arm, and thus will probably meet the hyperbola itself at four points.  So it is reasonable that this system of equations has four solutions.  (There are cases where there could be only two intersection points, but these must be carefully contrived.)
We need to be more analytical about this in order to better resolve the matter.  Rather than work with the equation of the hyperbola itself, we will use the equations of its asymptotes, found from
$$ (y \ - \ 1)^2 \ - \ x^2 \ = \ 1 \ \ \rightarrow \ \ (y \ - \ 1)^2 \ - \ x^2 \ = \ 0 \ \ \text{ [in the limit] }  $$
$$ \Rightarrow \ \ (y \ - \ 1)^2 \ = \ x^2 \ \ \Rightarrow \ \ y \ - \ 1 \ = \ \pm x \ \ \Rightarrow \ \ y \ = \ 1 \ \pm \ x \ \ .   $$
We will examine the intersection points of the parabola with these asymptotes by inserting their equations into that of the parabola, which produces two quadratic equations:
$$ ( \ 1 \ + \ x \ )^2 \ = \ -2x \ + \ 6 \ \ \Rightarrow \ \ x^2 \ + \ 4x \ - \ 5 \ = \ 0 \ \ ,  $$
$$ ( \ 1 \ - \ x \ )^2 \ = \ -2x \ + \ 6 \ \ \Rightarrow \ \ x^2 \ - \ 5 \ = \ 0 \ \ .  $$
We do not have to solve these for values of $ \ x \ $ (although it is quite easy to do so):  we only need note that both has positive discriminants, and thus each has two solutions for $ \ x \ $ .  Hence, there are four intersection points between the parabola and the asymptotes of the hyperbola, making it at least very probable that the system of equations has four ordered-pair solutions.  (Again, we could imagine a situation in which the parabola intersects the asymptotes, but just misses the hyperbola on one of its branches; this would again need to be delicately arranged.)
[It is readily found from the equations above that one of the asymptotes is intersected at $ \ x \ = \ \pm \sqrt{5} \ \approx \ \pm 2.24 \ $ , and the other at $ \ x \ = \ -5 \ $ and $ \ x \ = \ 1 \ $ . We see this in the graph below.]

It might also be mentioned that the intersection points of two conic sections also lie on a conic section.  If we set the equations of our two curves equal to one another, we obtain
$$ x^2 \ - \ y^2 \ + \ 2y \ = \ 2x \ + \ y^2 \ - \ 6 \ \ \Rightarrow \ \ x^2 \ - \ 2x \ - \ 2y^2 \ + \ 2y \ = \ -6 $$
$$ \Rightarrow \ \ ( x^2 \ - \ 2x \ + \ 1) - \ 2 \ (y^2 \ - \ y \ + \ \frac{1}{4}) \ = \ -6 \ + \ 1 \ - \ \frac{1}{2} $$
$$ \Rightarrow \ \ 2 \ (y \ - \ \frac{1}{2})^2 \ - \ (x \ - \ 1)^2 \ = \ \frac{11}{2} \ \ , $$
which is a "vertical" hyperbola with center $ \ ( 1, \ \frac{1}{2}) \ $ . (We could use the properties of this equation to show that there are four solutions, but the effort involved gets close to what would be required to solve it outright.) 
The graph below, which is based on the one Rnjai Lamba has posted, shows all of the curves we have discussed.

The parabola is marked in blue, the hyperbola in dark green (with its asymptotes in lighter green), and their intersection points and the hyperbola they lie on is in red.
A: Yes that is a good idea,this is how it will look.
[Courtesy:Wolfram Alpha]

