Write the equation in standard form and classify the conic section it defines.
4x^2-y^2+56x+4y+188=0

- anonymous

Write the equation in standard form and classify the conic section it defines.
4x^2-y^2+56x+4y+188=0

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- schrodinger

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- anonymous

@dan815

- anonymous

@Hero
@thesmartone
@zepdrix

- zepdrix

\[\large\rm 4x^2-y^2+56x+4y+188=0\]Mmm so I guess we'll need to complete some squares :)

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## More answers

- zepdrix

Let's group some stuff together,\[\large\rm \color{royalblue}{4x^2+56x}\color{orangered}{-y^2+4y}=-188\]Do you understand that step ok?

- anonymous

well they gave me choices but i cant copy or paste it. pretty much have to determine what type of conic section it is

- anonymous

i get that you have to put everyhting on the same side

- zepdrix

nah the 188 is fine over there for now :)
we gotta complete some squares though.
So for the blue stuff, factor a 4 out of each term,\[\large\rm \color{royalblue}{4(x^2+14x)}\color{orangered}{-y^2+4y}=-188\]

- zepdrix

Do you remember how to complete the square?
you take half of your b coefficient, and square it.
\[\large\rm \left(\frac{\color{royalblue}{14}}{2}\right)^2=49\]This is the value that completes the square for us.

- anonymous

wait a min this looks different. Where did you get the 14 from?

- zepdrix

The `x stuff` was,\[\large\rm 4x^2+56x\]Factoring a 4 out of each term gave me,\[\Large\rm 4(x^2+14x)\]56 divided by 4 is 14.

- anonymous

why would you factor that out?

- zepdrix

To try and put the equation into standard form.
We don't want a number attached to the squared x term.

- anonymous

ohhhh ok

- zepdrix

\[\Large\rm 4(x^2+14x+49)\]We would like to add 49 like this, to complete our square.
But we have to keep things balanced, we'll have to subtract 49 as well.
Or we can add 49 to each side if you're more comfortable with that.
But wait! It's not really a 49! Notice that it's inside of the brackets.
It's really the 4 TIMES 49.
So that's what needs to be given to each side,\[\large\rm \color{royalblue}{4(x^2+14x+49)}\color{orangered}{-y^2+4y}=-188+49\cdot4\]And then the blue will condense down,\[\large\rm \color{royalblue}{4(x+7)^2}\color{orangered}{-y^2+4y}=-188+49\cdot4\]x plus half of the b coefficient that we started with.
Take a moment and lemme know if that's confusing +_+
Maybe we can try some shortcuts if this process isn't working for you

- anonymous

im sorry im lost

- zepdrix

Maybe we can cheat a little bit then.
When we're given a big mess like this:\[\large\rm 4x^2-y^2+56x+4y+188=0\]We really only need to pay attention to the square terms, those are what tell us about the shape of the function.
We'll have a circle when both x^2 and y^2 are `positive` and have the `same` number attached. Example:\[\Large\rm 4x^2+4y^2-16x+8y-12=0\]See how both x^2 and y^2 are positive, and they both have the same number, a 4?

- anonymous

yes i see it

- zepdrix

We'll have an `ellipse` when both x^2 and y^2 are positive,
and they have `different` numbers attached to them.
Example:\[\Large\rm 4x^2+7y^2+12x-4y-12=0\]Notice that here, the 4 and 7 do not match, so we're ending up with an ellipse.
We don't care about the other numbers, those tell us something else, but not about the shape.

- zepdrix

We'll have a `hyperbola` when either x^2 OR y^2 is negative.
Example:\[\Large\rm 4x^2-4y^2+4x-3y-12=0\]Notice that for the hyperbola, it doesn't matter whether or not the numbers match.
We just care that one of the squared terms is negative. Only one of them.

- zepdrix

So for our problem,\[\large\rm \color{royalblue}{4x^2-y^2}+56x+4y+188=0\]only pay attention to the squares.
Which shape do you think we end up with?

- anonymous

circle?

- zepdrix

So for circle,
they both must be positive, oops our equation fails that already! the y^2 is negative.
and also, the numbers must be the same, oops the x^2 has a 4, the y^2 does not.

- anonymous

that last part triped me up

- zepdrix

one of them is negative,
so it looks like we're going to end up with a `hyperbola`, ya? :)
Does that give you an answer choice that you can go with?
Or do we need to narrow it down further because of the standard form?

- anonymous

no i think i got it

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