Did I do this correctly? And what conic does the equation represent?
(2/5x^2)-2y+4x+3y^2-4=0
(2/5x^2+ 4x +__) + (3y^2 -2y+__)= 4+__ + __
2/5(x^2 + 10x +25) + 3(y^2 - 2/3y+ 1/9)= 4 + 10 + 1/3
2/5(x+5)^2 + 3(y-1/3)^2 =43/3
[6(x+5)^2]/215 + [9(y-1/3)^2]/43 = 1
Conic: I think ellipse
If I did this incorrectly please correct my mistake thank you!
I got this from wikipedia.
In the Cartesian coordinate system, the graph of a quadratic equation in two variables is always a conic section, and all conic sections arise in this way. The equation will be of the form
Ax^2 + Bxy + Cy^2+ Dx + Ey + F = 0 with A, B, C not all zero. then:
Note that A and B are just polynomial coefficients, not the lengths of semi-major/minor axis as defined in the following sections.
- if B^2 − 4AC < 0, the equation represents an ellipse (unless the conic is degenerate, for example x^2 + y^2 + 10 = 0);
- if A = C and B = 0, the equation represents a circle;
- if B^2 − 4AC = 0, the equation represents a parabola;
- if B^2 − 4AC > 0, the equation represents a hyperbola;
- if we also have A + C = 0, the equation represents a rectangular hyperbola.