Converting from General to Standard

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The discussion revolves around converting the equation 2x^2 - y^2 - 4x - 8 = 0 into standard form. The initial steps involve rearranging and completing the square, leading to the conclusion that the equation represents a hyperbola. The correct standard form is identified as (x-1)²/(√10/√2)² - y²/(√10)² = 1. Participants clarify the missteps in calculations, particularly in handling constants during the transformation. The conversation emphasizes the importance of accurate algebraic manipulation in deriving the correct standard form.
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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
 
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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.
 
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dextercioby said:
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.

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?
 
dextercioby said:
When adding & subtracting 2.

Daniel.
:confused:
 
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.
 
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