anonymous
  • anonymous
Write the equation in standard form and classify the conic section it defines. 4x^2-y^2+56x+4y+188=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@Hero @thesmartone @zepdrix
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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zepdrix
  • 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
  • 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
  • anonymous
i get that you have to put everyhting on the same side
zepdrix
  • 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
  • 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
  • anonymous
wait a min this looks different. Where did you get the 14 from?
zepdrix
  • 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
  • anonymous
why would you factor that out?
zepdrix
  • zepdrix
To try and put the equation into standard form. We don't want a number attached to the squared x term.
anonymous
  • anonymous
ohhhh ok
zepdrix
  • 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
  • anonymous
im sorry im lost
zepdrix
  • 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
  • anonymous
yes i see it
zepdrix
  • 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
  • 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
  • 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
  • anonymous
circle?
zepdrix
  • 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
  • anonymous
that last part triped me up
zepdrix
  • 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
  • anonymous
no i think i got it

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