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Homework Help: Standard form of an ellipse

  1. Mar 26, 2005 #1
    When in standard form how do you know whether the ellipse is horizontal or vertical? and how do you know what a= and b= for the major and minor axis, and vertices? How do you get the coordinates of the foci?
  2. jcsd
  3. Mar 26, 2005 #2
    If the denominator attached to the x is less than that of the y then it will be vertical, (as there is less of a distance between the x intercepts that the y).

    The semi-major axis will be the root of the largests denominator
    The semi-minor axis will be the root of the smallest denominator
    The coordinates of the center will be (h,k) for and elispse in the form (just a simple translation)

    [tex]\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}=1[/tex]
  4. Mar 26, 2005 #3
    how about the coordinates of the foci?
  5. Mar 26, 2005 #4


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    If you know what's special about the covertex, there's a very easy way to find the coordinates of each focus.

    Both focii lie on the major axis. For a horizontal ellipse, your coordinates have to be (x,0). For a vertical ellipse, (0,y).

    The covertex is the point where the minor axis intersects the ellipse. Half of the minor axis is the semi-minor axis (b) and half of the major axis is the semi-major axis (a). The distance from either focus to the covertex is equal to the semi-major axis.

    Now you know two sides of your triangle. You know the hypotenuse which happens to have a length equal to the semi-major axis (a). You know the semi-minor axis (b) which forms one leg of your triangle. You also know the minor axis is perpendicular to the major axis, so you know you have a right triangle.

    You use the Pythagorean thereom to find the missing length: the length from your origin to the focus. The length will be equal to your linear eccentricity (c). In other words, you have [tex]a^2=b^2+c^2[/tex]. For a horizontal ellipse, the length is the missing x variable in your coordinates. For a vertical ellipse, the missing y variable. For the opposite focii, just reverse the sign (positive x to negative x, pos y to neg y, as applicable).
  6. Mar 26, 2005 #5
    lol woh bob u got way too complicated how did triangles get into this :rofl:
  7. Mar 26, 2005 #6


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    Maybe a drawing would help. I drew a horizontal ellipse, where the major axis lies along the x-axis. The coordinates of your focii would be (c,0) and (-c,0)

    Attached Files:

  8. Mar 26, 2005 #7
    how do I open the attachment?
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