- #1

- 86

- 0

x^2-2xy+y^2+5x+5y=0

using the formulas b^2-4ac=0 (indicates a parabola)

b^2-4ac<0 (indicates an ellipse)

b^2-4ac>0 (indicates a hyperbola)

2^2-4(1)(1)=0

4-4=0 therefore this graph must be a parabola!

Am I correct?

- Thread starter TonyC
- Start date

- #1

- 86

- 0

x^2-2xy+y^2+5x+5y=0

using the formulas b^2-4ac=0 (indicates a parabola)

b^2-4ac<0 (indicates an ellipse)

b^2-4ac>0 (indicates a hyperbola)

2^2-4(1)(1)=0

4-4=0 therefore this graph must be a parabola!

Am I correct?

- #2

TD

Homework Helper

- 1,022

- 0

It's indeed a parabola

- #3

- 86

- 0

YIPPEE, thank

- #4

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 132

[tex](x-y)^{2}+5(x+y)=0[/tex]

This can be brought onto the form:

[tex]u=-\frac{\sqrt{2}}{5}v^{2}, u=\frac{x+y}{\sqrt{2}}, v=\frac{x-y}{\sqrt{2}}[/tex]

where the u-v axes are 45 degrees rotated with respect to the xy axes.

- Last Post

- Replies
- 7

- Views
- 1K

- Last Post

- Replies
- 5

- Views
- 786

- Replies
- 3

- Views
- 814

- Last Post

- Replies
- 1

- Views
- 917

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 6K

- Replies
- 9

- Views
- 3K

- Last Post

- Replies
- 4

- Views
- 689

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 7

- Views
- 3K