Understanding Ellipse Standard Form and Axis Orientation

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Homework Help Overview

The discussion revolves around understanding the standard form of ellipses, specifically how to determine their orientation (horizontal or vertical), identify the lengths of the semi-major and semi-minor axes, and calculate the coordinates of the foci.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore how the denominators in the standard form indicate the orientation of the ellipse. Questions arise regarding the determination of the semi-major and semi-minor axes, as well as the coordinates of the foci. Some participants discuss the relationship between the axes and the covertex, and how to apply the Pythagorean theorem to find the foci.

Discussion Status

There is an ongoing exploration of the relationships between the components of the ellipse. Some participants provide insights into the calculations involved, while others express confusion or seek clarification on the geometric interpretations presented.

Contextual Notes

Participants are discussing the implications of the standard form of an ellipse and its geometric properties, including the use of the Pythagorean theorem in relation to the foci. There is a mention of a drawing that may aid in understanding, indicating a visual component to the discussion.

aisha
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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?
 
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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]
 
how about the coordinates of the foci?
 
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).
 
lol woh bob u got way too complicated how did triangles get into this :smile:
 
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)
 

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how do I open the attachment?
 

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