Write an quation for each parabola with vertex at orgin.
vertex (2,1), focus (-2,-3)
From the given points you can calculate the equation of the axis, the directrix d and the coordinates of the points B and C.
EDIT: If you have the general equation of a conic:
Ax² + 2Bxy + Cy² + 2Dx + 2Ey + F = 0
then you can determine which conic you have by calculating
AC - B² > 0 ----> ellipse
AC - B² = 0 ----> parabola
AC - B² < 0 ----> hyperbola
You'll see that the coefficients of the equation I posted satisfy the condition indicating a parabola.
I haven't worked that much with transformations of conics. I knew the axis of symmetry was oblique, but had forgotten about identifying the conic by using the discriminant.
So this parabola is a rotation about the origin, and the angle of rotation is pi/2 (45 degrees) since A = C.
I am more familiar with: Ax^2 + Bxy + Cy^2 + Dx +Ey + F = 0
as the general equation. Then, testing the discriminant (B^2 - 4AC),
If B^2 - 4AC < 0, and A = C, then it's a circle. If A not equal C, then it's an ellipse
If B^2 -4AC > 0, then it's a hyperbola
If B^2 - 4AC = 0, then it's a parabola.
I have to think the OP made a typo on this one, but it turned out to be a great post....for me, anyway.