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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Hero @thesmartone @zepdrix

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm 4x^2y^2+56x+4y+188=0\]Mmm so I guess we'll need to complete some squares :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Let'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
 one year ago
Best ResponseYou've already chosen the best response.0well they gave me choices but i cant copy or paste it. pretty much have to determine what type of conic section it is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i get that you have to put everyhting on the same side

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1nah 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
 one year ago
Best ResponseYou've already chosen the best response.1Do 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
 one year ago
Best ResponseYou've already chosen the best response.0wait a min this looks different. Where did you get the 14 from?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1The `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
 one year ago
Best ResponseYou've already chosen the best response.0why would you factor that out?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1To try and put the equation into standard form. We don't want a number attached to the squared x term.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large\rm 4(x^2+14x+49)\]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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Maybe we can cheat a little bit then. When we're given a big mess like this:\[\large\rm 4x^2y^2+56x+4y+188=0\]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:\[\Large\rm 4x^2+4y^216x+8y12=0\]See how both x^2 and y^2 are positive, and they both have the same number, a 4?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1We'll have an `ellipse` when both x^2 and y^2 are positive, and they have `different` numbers attached to them. Example:\[\Large\rm 4x^2+7y^2+12x4y12=0\]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
 one year ago
Best ResponseYou've already chosen the best response.1We'll have a `hyperbola` when either x^2 OR y^2 is negative. Example:\[\Large\rm 4x^24y^2+4x3y12=0\]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
 one year ago
Best ResponseYou've already chosen the best response.1So for our problem,\[\large\rm \color{royalblue}{4x^2y^2}+56x+4y+188=0\]only pay attention to the squares. Which shape do you think we end up with?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.0that last part triped me up

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1one 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?
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