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What is the difference in these two equations?

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the difference in these equations?

    1. 4(x-1)^2)+(y+2)^2=0

    2. (((x-1)^2)/4)+(((y+2)^2)/16)=1

    3. ((-2(x-1)^2)/1)-(((y+2)^2)/2)=1

    2. Relevant equations

    -None-

    3. The attempt at a solution

    If you are wondering where i got these equations from where by changing the constant to +8, and +10 from -8 for this equation:

    4x^2+y^2-8x+4y-8=0

    I think the first one is a circle because it kind of looks like x^2+y^2=r^2, not sure..
    The second one is just a basic ellipse because it follows its standard form.
    The third one is a hyperbola because it is x^2-y^2
     
  2. jcsd
  3. Mar 14, 2012 #2

    Mark44

    Staff: Mentor

    Nope, it's not a circle. If it were 4x2 + y2 = 1, what would it be?

    How about if it were 4x2 + y2 = 0?
    Yes.
    No it isn't. And you don't have x2 - y2 = 1, which is a hyperbola. Look closer at the signs.
     
  4. Mar 14, 2012 #3
    That means the first one is a 'v' graph? Do they have a certain name?

    And the third one is a parabola.
     
  5. Mar 14, 2012 #4

    Mark44

    Staff: Mentor

    No. Think about what the graph of 4(x-1)^2+(y+2)^2=0 looks like, and in particular, think about what happens to a number when you square it. The graph is very simple.

    No again. In the equation of a parabola there will be an x2 term and no y2 term, OR a y2 term but no x2 term. In this equation --
    ((-2(x-1)^2)/1)-(((y+2)^2)/2)=1

    -- there are both an x2 term and a y2 term. Simplify the equation a bit and see if you can graph it.
     
  6. Mar 14, 2012 #5
    Didn't you simplify the first one already, because the x-1 and y+2 are just origin shifts so it would look like:

    4x^2+y^2=0

    I am thinking parabola...

    The second one would be:

    (-2x^2)-(y^2/2)=1
     
  7. Mar 14, 2012 #6

    Mark44

    Staff: Mentor

    If you translate this graph 1 unit right and 2 units down, you get the graph of your first equation.
    No, it can't be a parabola, because there is an x2 term AND a y2 term. Parabolas never have two second-degree terms.

    If the equation were 4x2 + y2 = 1, what would it be?

    What would the untranslated equation 4x2 + y2 = 0 be? What can you put in for x and y so that you get 0 on the right side?
    Well, all you did here was remove the translations. There's one more simplification that you can do.
     
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