Standard form of an ellipse

1. Mar 26, 2005

aisha

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. Mar 26, 2005

mtong

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)

$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}=1$$

3. Mar 26, 2005

aisha

how about the coordinates of the foci?

4. Mar 26, 2005

BobG

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 $$a^2=b^2+c^2$$. 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).

5. Mar 26, 2005

aisha

lol woh bob u got way too complicated how did triangles get into this :rofl:

6. Mar 26, 2005

BobG

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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7. Mar 26, 2005

aisha

how do I open the attachment?