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Find the equation of an ellipse that has its center at (5,1) and foci at (5,4) and (5,-2)

Any kind of help will be much appreciated thank you everyone

2006-12-06 15:54:27 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

I've been thinking about this for half an hour, and I hate to say this, but I think you don't have enough information.

The definition of an ellipse is "the set of all points (x,y) in a plane whose distance to two points (the foci) is a constant".

In fact, the constant is 2a, where a is the "semi-major axis".

The basic equation of an ellipse is:

(x - h)/a² + (y - k)/b² = 1

Where (h, k) is the center and a and b are the lengths of the major and minor axes.

The other equation of interest in for ellipses is:

a² + b² = c²

Where a and b are the same as above, and c is the "focal length" or distance from the foci to the center.

You're given the center and the foci. The foci have different y's, so this means that the major axis will be the y - axis.

Also the focal length, c, is the difference between the y-coordinates of the center and either of the foci: 4 -1 = 3, or 1 - (-2) = 3.

So we have 2 equations now:

(x - 5)²/a² + (y - 1)²/b² = 1

and

a² + b² = (3)² = 9

But the problem here is that you have 2 equations with 4 unknowns. You don't have a point on the ellipse, you don't have the eccentricity of the ellipse...you don't have any way of figuring out exactly what a and b are.

Another way to think about it:

One way you can draw an ellipse is to push two pins into the paper at the foci, tie a string to each, pull it taut, and draw.

In this case, we know where the pins go, but we're given no way to figure out how long the string is.

Did you leave out part of the question maybe?

2006-12-06 16:28:56 · answer #1 · answered by Jim Burnell 6 · 0 0

It is true that you do not have enough information. One additional number is needed. You do have the fact that the foci are 6 units apart along the y-axis. If you call the x semiaxis a, and the y-semiaxis n and th ecenter of the ellipse (xc, yc) then the equaitons is:

((x-xc)/a)^2 + ((y-yc)/b)^2 = 1

For your ellipse there are 4 unknowns, xc, yc, a, and b. You have specified xc, yc, and one more parameter, the distance from the center to a focus (c). This uses the additional relation:

a^2 = b^2 + c^2

So it gives some information about a and b, but you still are one variable short.

2006-12-07 01:23:55 · answer #2 · answered by Pretzels 5 · 0 0

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