Is anyone good with hyperbolas? If so, I would really appreciate the help.
There's four more.
you got the center?
The center is (-1,3) right?
you need a couple facts first you need the center then you need to know whether the \(y^2\) term comes first, or whether the \(x^2\) term comes first the center you should see with your eyeballs you see it?
no, you have the \(x\) and \(y\) backwards
right 3, down 1, center is \((3,-1)\) not \((-1,3)\)
Oh. ;o oops. (3, -1)
ok so using that fact we know at least one thing there will be a \((x-3)^2\) term and a \((y+1)^2\) term is that part clear?
Yes.
ok next question is it going to be \[\frac{(x-3)^2}{a^2}-\frac{(y+1)^2}{b^2}=1\] or is it going to be \[\frac{(y+1)^2}{b^2}-\frac{(x-3)^2}{a^2}=1\] i.e. which comes first you know that from the graph as well do you know which?
Uh... Wouldn't it be the first one? Because x goes before y...
no it is not always the \(x\) first, that is for a coordinate
|dw:1399260766344:dw|
|dw:1399260790957:dw|
clear?
Oh so it's y... Y is first... Right?
right
now we don't really have to worry about the denominators, although we could figure them out but you only have one choice that fits the bill we need
Oh wait... I got 'em backwards again... It's C.... Correct?
D* oy... I'm confusing myself... It would be D, because the -3 is with the x, right?
hold on let me look again
yeah D
c: Thank yhu. Can yhu help me with the others? I'm not understanding this unit, and my notes are confusing... And not very helpful.
sure are they going to be like that one?
Sort of yes.
k
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