In fact, a conic has 4 foci. We can see this if we look at a canonical ellipse,which is wide and short, and start making it smaller in the direction of the x-axis. The two foci get closer, until we reach a circle when they collapse to one point. Then, if we continue they start to have a different trajectory - up and down. This cannot be, and what really happens is that two foci escape to the complex part of the plane, while the other two arrive from it.
A purely geometro algebraic definition of foci (well not exactly, because it depends on the projective coordinates you chose!) is the followin:
take the points I=[1:i:0] and J=[1:-i:0], and a conic C. By bezout's theorem there are two lines through I that are tangent to C and likewise with J. The two pair of lines intersect at 4 points, which are precisely the foci.
Remark: The fact that there are two tangents from a point to C does not follow immediately from Bezout's, "tangency" is not a linear condition. However, it does follow if first we use projective duality with respect to C.
What is the definition of a projective change of coordinates in $ℝℙ^2$? Is it the same as the definition for $ℂℙ^2$ but with all real coefficients?
Well, what is your definition of that “change of coordinates” in $ℂℙ^2$? Usually the term I'd prefer would be “change of basis”. A projective basis in the projective plane can be given by four points, and with respect to that basis all points in the plane can be expressed using homogeneous coordinate vectors. The convention is to associate the four points with the homogeneous coordinates $[1,0,0]$, $[0,1,0]$, $[0,0,1]$ and $[1,1,1]$.
If this is what you have in mind, then a change of basis would be the application of a projective transformation, i.e. the multiplication of point vectors with an invertible $3\times3$ matrix over the base field, be it $\mathbb C$, $\mathbb R$ or something else.
While you just multiply such a matrix with the vector of a point, a conic as a quadratic form is usually described by a symmetric matrix, and you have to apply the transformation matrix from both sides. To be more specific, assume that $p$ is on the original conic $C$. That means $p^T\cdot C\cdot p=0$. The transformed point $p'=M\cdot p$ lies on the transformed conic $C'$ if $p'^T\cdot C'\cdot p'=0$. Which you can achieve by taking $C'=(M^{-1})^T\cdot C\cdot(M^{-1})$ since that way the matrices $M$ and $M^{-1}$ will cancel out.
So you can say that two conics $A$ and $B$ are projectively equivalent if there exists a matrix $M$ such that $M^T\cdot A\cdot M=B$.
Now consider the eigendecomposition of such a conic matrix $A$. A real symmetric matrix is always decomposable [1], so you always have $A=Q\cdot\Lambda\cdot Q^T$ with $\Lambda$ diagonal. If you interpret $Q^T$ as your projective transformation, that means every real conic is projectively equivalent to one which can be expressed by a diagonal matrix.
In the next step, you can apply a scaling transformation to simplify this further. For example, assume that $\lambda_1$ is the first eigenvalue, the first diagonal element of $\Lambda$. Then you can multiply with a diagonal matrix which has $1/\sqrt{\lvert\lambda_1\rvert}$ as its first entry. That way you can bring that first entry to $\pm1$. Except if it is equal to zero, in which case scaling it to magnitude one won't work. Do this scaling for all three elements and you end up with a matrix with has diagonal elements taken from $\{-1,0,1\}$.
Now you can do some permutations of coordinates, to rearrange elements. And you can also replace $A$ by $-A$ and still describe the same conic. So at the end of the day, you can say that every real conic is up to projective transformations equivalent to one where the diagonal elements are one of the following:
- $(1,1,-1)$, e.g. $x^2+y^2-1=0$. That's your regular real non-degenerate conic.
- $(1,1,1)$, e.g. $x^2+y^2+1=0$. That's a non-degenerate conic which contains no real points. Purely complex despite the real coefficients.
- $(1,-1,0)$, e.g. $(x+y)(x-y)=0$. A pair of real lines.
- $(1,1,0)$, e.g. $x^2+y^2=0$. A pair of conjugate complex lines, with their real point of intersection as the only real point in the conic.
- $(1,0,0)$, e.g. $x^2=0$. A single real line with multiplicity 2.
Are all non-degenerate conic sections (i.e. ellipses, hyperbolas, and parabolas), still equivalent in $ℝℙ^2$, the same way they are equivalent in $ℂℙ^2$?
All non-degenerate conics which do contain real points are projectively equivalent. Ellipses, hyperbolas and parabolas do fall into this category. However there is also a class of quadratic forms which has no real solutions, and if you consider these as conics as well, then no, they are not equivalent to the former.
Over the complex numbers, the above classification would become even simpler, because multiplication by $i$ from both sides will convert between $-1$ and $+1$. So you'd only distinguish by how many zeros there are. On the other hand, the statement that every symmetric matrix is diagonalizable holds only over the reals, so that part of the argument may need more work.
To gain some intuition as to why ellipses, parabolas and hyperbolas are the same, here is something in the spirit of this answer of mine. Imagine a conic, e.g. an ellipse. You can view a projective transformation as one of two things: either a transformation which takes your ellipse somewhere else, or as the inverse transformation applied not to the ellipse but to the reference frame, and to the line at infinity in particular. If you move the line at infinity so that in its new position it intersects the ellipse, then you've essentially turned the ellipse into a hyperbola. A hyperbola is just a non-degenerate conic which intersects the line at infinity in two distinct points. Between ellipse and hyperbola there is the parabola, where the line at infinity merely touches the conic in a single point. It's also easy to imagine a projective transformation taking the line at infinity to such a tangential position.
If you are restricted to affine transformations, then the line at infinity stays where it is, and so does the kind of conic you have. So the key point in the question What shape… is that it only considers affine transformations. The matrix $Q$ used in the diagonalization above will in general not be affine.
Best Answer
I'd not put it like this, even though you are essentially right. Instead I'd say that a generic projective transformation alters the line at infinity.
So let's say you have a hyperbola, which intersects the line at infinity in two points. Then you apply a projective transformation to map that to the unit circle. The image of the line at infinity under that transformation will be a finite line which intersects the unit circle in two points, namely the images of the original points of intersection. The new line at infinity, after the transformation was applied, has no real points in common with the unit circle. (Now the points of intersection would be the complex ideal circle points $[1:\pm i:0]$, but you don't need this for your question I guess.)