# Converting from General to Standard

1. Mar 14, 2005

### trigger352

I have the answer to this equation in standard form but I don't know how I they got it,

$$2x^2 - y^2 - 4x - 8 = 0$$

so far I get this:

$$2x^2 - 4x - y^2 - 8 = 0$$

$$2 (x^2 - 2x +1) - y^2 -1 - 8 = 0$$

$$2 (x-1)^2 - y^2 -1 - 8 = 0$$

the equation in standard form is...

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

Last edited: Mar 14, 2005
2. Mar 14, 2005

### dextercioby

It should be

$$\frac{(x-1)^{2}}{\left (\frac{\sqrt{10}}{\sqrt{2}}\right)^{2}}-\frac{y^{2}}{(\sqrt{10})^{2}}=1$$

A hyperbola...

Daniel.

Last edited: Mar 14, 2005
3. Mar 14, 2005

### trigger352

But why? How did you get that answer. The answer I got was from my teacher so I'm surprised he'd be wrong.

So where did I go wrong in my calculations?

4. Mar 14, 2005

### dextercioby

Daniel.

5. Mar 14, 2005

### trigger352

6. Mar 14, 2005

### dextercioby

$$2x^{2}-4x+2-2-y^{2}-8=0$$

$$2(x-1)^{2}-y^{2}=10$$

Then it's simple to reach to my formula.

Daniel.