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Write equation of translated graph in general form

  1. Jul 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Identify the graph of each equation. Write the equation of the translated graph for T(-5,6) in general form. Then draw the graph.

    4x2 + 5y2 = 20

    2. Relevant equations
    Ax2 + Bxy + Cy2 + Dx + Ey + F = 0


    3. The attempt at a solution
    I put the equation instandard form and got :

    [tex]\frac{x^2}{5}[/tex] + [tex]\frac{y^2}{4}[/tex] = 1

    This shows that it is a graph of an ellipse with center (0,0).

    However, I wasnt sure what to do next. I found the translated points using matrices and adding, but I dont know what to do with this, ot even if I need this information.
     
  2. jcsd
  3. Jul 9, 2010 #2
    Is it standard form?
    let's see,
    (standard form) = 0
    and your equation is
    (some equation) = 1
     
  4. Jul 9, 2010 #3

    Mark44

    Staff: Mentor

    If the graph of the ellipse is translated left by 5 units and up 6 units, the center will move from (0, 0) to (-5, 6). How will this affect the equation
    [tex]\frac{x^2}{5} + \frac{y^2}{4} = 1[/tex]?

    Tip: write your entire equation in one pair of [ tex] tags.
     
  5. Jul 9, 2010 #4
    The equation will then become:

    [tex]\frac{(x+5)^2}{5} + \frac{(y-6)^2}{4} = 1[/tex]

    NOw, how can I convert this into general form?
     
  6. Jul 9, 2010 #5

    Mark44

    Staff: Mentor

    Multiply both sides by 20, expand (x + 5) and (y - 6)2, and bring everything to one side, with zero on the other side.
     
  7. Jul 9, 2010 #6
    Ok, I get the following:


    [tex]4x^2 + 5y^2 + 40x - 60y +260 = 0[/tex]

    This looks good, except there is no "Bxy" term. When would this term appear?

    Looking through my older notes, I found a few problems where i converted from standard form to general form before. However, it looks like that way I did it was by setting up a system of 3 equations, and somehow solving for the variables D, E, and F. Can this method I mentioned by used on anything other than circles?


    Lastly, will the method you showed me work with all types of conic sections?
     
  8. Jul 9, 2010 #7

    Mark44

    Staff: Mentor

    The xy term shows up when conic sections (such as an ellipse in this problem) are rotated.
    I don't understand what you're saying here. To convert from general form to standard form (if I remember these terms correctly), you complete the square in the x terms and in the y terms. If you're talking about the general form with an xy term, it's been a very long time since I've done that, so can't commit on the method you describe.
     
  9. Jul 10, 2010 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If you have a conic section with [itex]Ax^2+ Bxy+ Cy^2[/itex] (and other non-squared terms) with C not 0, you can get rid of it in either of two ways:

    1) Let [itex]x= x' cos(\theta)- y' sin(\theta)[/itex], [itex]y= x' sin(\theta)+ y' cos(\theta)[/itex], replace x and y in your formula by that, and solve for a value of [itex]\theta[/itex] that make Bxy= 0. [itex]\theta[/itex] is the angle through which the axes are rotated and the entire equation with x', y' instead of x, y will have no "xy" term.


    2) Find the eigenvalues and eigenvectors of the matrix
    [tex]\begin{bmatrix}A & \frac{1}{2}C \\ \frac{1}{2}C & D\end{bmatrix}[/tex]

    The eigenvectors will point along line y= ax and y= bx, that are parallel to the axes of the conic and substituting x'= ax- y, y'= bx- y will get rid of the "xy" term.
     
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