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How to graph conic?

  1. Mar 28, 2005 #1
    [tex] 4(x+1)^2 - \frac {(y+2)^2} {9} = -1 [/tex]

    I think this is the equation for a hyperbola, I dont know how to graph this and state the key features can someone please help me? :redface:
     
  2. jcsd
  3. Mar 29, 2005 #2
    Try to isolate either x or y. That way, it will be easy for you to find the plotted points.

    If, however, you have a program, like Geometer's Sketchpad, you can plot that graph without doing anything.
     
  4. Mar 29, 2005 #3
    Quit flooding the board with repetitive questions, AISHA!
     
  5. Mar 29, 2005 #4

    HallsofIvy

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    You are going a little overboard here, Aisha.

    Do you know what the "standard" formula for ellipse, parabola, hyperbola look like? That should be in your textbook and your textbook should be your first resource.

    Yes, this is a hyperbola- one of the first things you should have learned is that two squares with the same sign is an ellipse (considering a circle as a special case of an ellipse), two square with different signs is a hyperbola, one variable squared but not the other is a parabola (all assuming there is no "xy" term- you really don't want to deal with that!).

    The standard form for a hyperbola is [tex]\frac{(x-a)^2}{u^2}- \frac{(y -b)^2}{v^2}= 1[/tex] or [tex]\frac{-(x-a)^2}{u^2}+ \frac{(y -b)^2}{v^2}= 1[/tex].

    The form you have is almost that: you will need to multiply the equation by -1. Also that [tex]4(x+1)^2[/tex] should be changed to [tex]\frac{(x+1)^2}{(\frac{1/2})^2}[/tex].

    The center of such a hyperbola is at (a,b) (you knew that didn't you?). A good way to graph it is this- from (a,b) measure distance u to right and left and v up and down so you can draw the rectangle with center at (a,b), sides of length 2u and 2d.
    Draw the diagonals. Those will be the asymptotes of the hyperbola, with equations
    y= (v/u)(x- a)+ b and y= (-v/u)(x-a)+b. To determine the vertices, you will need to decide whether the axis of the hyperbola is vertical (x= a) or horizontal (y= b).
    If x= a is the equation solvable for y? If so the axis is vertical (and the vertices are is at (a,b+v) and (a,b-v)), if not, the axis is horizontal (and the vertices are at (a+u,v) and (a-u,b)).
     
  6. Mar 29, 2005 #5
    If your teacher allows, then download graphmatica. Its easy to use, and also freeware. it allows you to type in any equation and it produces the graph from straight lines to tricky conics to trigonometry.
     
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