# Sketching level curves

1. Jan 29, 2009

### Somefantastik

$$x^{2}-y^{2}-2x+4y+5;$$

let $$x^{2}-y^{2}-2x+4y+5 \ = \ c;$$

To sketch this as a level curve, I'm not sure how to proceed. I can't seem to rearrange the function into anything familiar.

For the sake of trying to find a reference point, I let x=0 and found

$$y \ = \ 2 \ ^{+}_{-}\sqrt{9-c};$$

then y=0 =>

$$x \ = \ 1 \ ^{+}_{-}\sqrt{-4+c};$$

If I let c = 5, I get

$$x^{2}-y^{2}-2x+4y = 0;$$

which gives x = 0, x = 2, y = 0, y = 4.

What should I do next?

2. Jan 29, 2009

### NoMoreExams

Level curves are usually for functions f(x,y) = blah i.e. when you have 2 ind. variables. Here you only have one ind.... so you can solve for y by completing the square

3. Jan 29, 2009

### Somefantastik

The function that I am trying to draw level curves for is

$$V(x,y) =x^{2}-y^{2}-2x+4y+5;$$

4. Jan 29, 2009

### NoMoreExams

Well you should be able to rewrite it by completing the square, I'll work this one as example

$$V(x,y) = x^2 - y^2 - 2x + 4y + 5 = (x-1)^2 - (y-2)^2 + 8$$

5. Jan 30, 2009

### Somefantastik

Ok, yeah I see that. It's a translated hyperbola?

I keep getting

$$V(x,y) = (x-1)^{2} - (y-2)^{2} +5;$$

Which is a hyperbola centered at (1,2) and translated somewhere 5 units?

6. Jan 30, 2009

### NoMoreExams

How are you getting that?

$$(x-1)^2 - (y-2)^2 + 5 = x^2 - 2x + 1 - (y^2 - 4y + 4) + 5 = x^2 - y^2 - 2x + 4y + 2$$

That's not what you started with?

7. Jan 30, 2009

### Somefantastik

I should have been paying better attention.

How would I graph this? My cal books just have all these guys equal to 1.

$$V(x,y) = (x-1)^2 \ - \ (y-2)^2 \ + \ 8 \ = \ c;$$
$$V(x,y) = (x-1)^2 \ - \ (y-2)^2 \ = \ c \ - 8;$$

8. Feb 1, 2009

Well, if $(x - 1)^2 - (y - 2)^2 = c - 8$, then
$$\frac{(x - 1)^2}{c - 8} - \frac{(y - 2)^2}{c - 8} = 1.$$