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### Graph the Ellipse...

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Date: 01/01/98 at 17:15:24
From: Callie White
Subject: PreCalculus

[((x-1)^2)/2^2] + [((y-2)^2)/2^2]=1

Graph the ellipse and find the major and minor axes, the foci, the
eccentricity, and the center.
```

```
Date: 01/06/98 at 16:31:36
From: Doctor Jaffee
Subject: Re: PreCalculus

Hi Callie,

I think your teacher might be trying to trick you with this one
(assuming that this is a question you brought from your class).

If you multiply both sides of the equation by 4 (that's the 2^2 in
your equation), the denominators will cancel and the resulting
equation will be (x-1)^2 + (y-2)^2 = 4. Any equation in the form
(x-h)^2 + (y-k)^2 = r^2 is a circle whose center is at the point
(h,k) and whose radius is r.

So, what you really have is a degenerate ellipse ("degenerate" is
really the technical term for a geometric figure that has changed in a
particular way to another, simpler form. I'm not trying to insult the
ellipse). At any rate, you have a circle whose center is at the point
(1,2) and whose radius is 2.

The two foci have become the same point, the center, and neither axis
through the center is major or minor. In fact, any diameter can be
considered one of the axes. So, you could consider the two axes to be
the one from (-1,2) to (3,2) and the other from (1,0) to (1,4).

And finally the question about the eccentricity. One definition of
eccentricity is the ratio of the distance from the center to a focus
and the distance from the center to an endpoint of the major axis.
Since the focus is at the center, the distance from the center to the
focus is zero, so the ratio is zero.  Thus, the eccentricity is 0.

I hope this has helped.

-Doctor Jaffee,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Equations, Graphs, Translations

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