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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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    @dan815

  2. anonymous
    • one year ago
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    @Hero @thesmartone @zepdrix

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

  4. zepdrix
    • one year ago
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    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?

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

  6. anonymous
    • one year ago
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    i get that you have to put everyhting on the same side

  7. zepdrix
    • one year ago
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    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,\[\large\rm \color{royalblue}{4(x^2+14x)}\color{orangered}{-y^2+4y}=-188\]

  8. zepdrix
    • one year ago
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    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.

  9. anonymous
    • one year ago
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    wait a min this looks different. Where did you get the 14 from?

  10. zepdrix
    • one year ago
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    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.

  11. anonymous
    • one year ago
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    why would you factor that out?

  12. zepdrix
    • one year ago
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    To try and put the equation into standard form. We don't want a number attached to the squared x term.

  13. anonymous
    • one year ago
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    ohhhh ok

  14. zepdrix
    • one year ago
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    \[\Large\rm 4(x^2+14x+49)\]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

  15. anonymous
    • one year ago
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    im sorry im lost

  16. zepdrix
    • one year ago
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    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?

  17. anonymous
    • one year ago
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    yes i see it

  18. zepdrix
    • one year ago
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    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.

  19. zepdrix
    • one year ago
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    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.

  20. zepdrix
    • one year ago
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    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?

  21. anonymous
    • one year ago
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    circle?

  22. zepdrix
    • one year ago
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    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.

  23. anonymous
    • one year ago
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    that last part triped me up

  24. zepdrix
    • one year ago
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    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?

  25. anonymous
    • one year ago
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    no i think i got it

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