Question 442912: Can you show a step by step process of turning x^2+y^2-6x-2y+6 into standard form for a hyperbola? Found 2 solutions by Alan3354, swincher4391:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Can you show a step by step process of turning x^2+y^2-6x-2y+6 into standard form for a hyperbola?
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It's not a hyperbola.
It's not an equation, there's no equal sign.
You can put this solution on YOUR website! It may be best to complete the square here:
Let's make a couple groupings here:
[x^2-6x] + [y^2 -2y] = -6
Let's complete the square with the x's.
Recall, to complete the square, take the and make that the c term.
So our c will be (-6/2)^2 = 9
[x^2-6x + 9] + [y^2 -2y] = -6 + 9
You must balance the equation by adding what you added to the xs.
Same process with the ys.
[x^2-6x+9] + [y^2 -2y + 1] = -6 + 9 + 1
The reason we did this was so we can now factor the xs and ys into a square.
[x^2-6x + 9] = (x-3)^2 ... it will always be
[y^2-2y + 1] = (y-1)^2
So now we have
To be in standard form, we must be in this form
We have a problem.
This is actually the equation for a circle.
tells us we are centered at (3,1) with a radius of 2.
Is it possible that either the or is supposed to be negative? Then you'd have yourself a hyperbola.
