# Math Help - completing the square

1. ## completing the square

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!

2. Originally Posted by yoman360
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'll rewrite the steps using LaTeX so I can have an easier time reading and checking...

$\frac{2}{5}x^2 - 2y + 4x + 3y^2 - 4 = 0$

$\frac{2}{5}x^2 + 4x + 3y^2 - 2y = 4$

$\frac{2}{5}(x^2 + 10x) + 3\left(y^2 - \frac{2}{3}y\right) = 4$

$\frac{2}{5}(x^2 + 10x + 5^2 - 5^2) + 3\left[y^2 - \frac{2}{3}y + \left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right)^2\right] = 4$

$\frac{2}{5}\left[(x + 5)^2 - 25\right] + 3\left[\left(y - \frac{1}{3}\right)^2 - \frac{1}{9}\right] = 4$

$\frac{2}{5}(x + 5)^2 - 10 + 3\left(y - \frac{1}{3}\right)^2 -\frac{1}{3} = 4$

$\frac{2(x + 5)^2}{5} + 3\left(y - \frac{1}{3}\right)^2 = \frac{43}{3}$

$\frac{3}{43}\left[\frac{2(x + 5)^2}{5} + 3\left(y - \frac{1}{3}\right)^2\right] = 1$

$\frac{6(x + 5)^2}{215} + \frac{9\left(y - \frac{1}{3}\right)^2}{43} = 1$.

I get the same answer as you , and yes, it's an ellipse.

3. Originally Posted by yoman360
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:
• 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;
Note that A and B are just polynomial coefficients, not the lengths of semi-major/minor axis as defined in the following sections.