PreCalc - Ellipses


by Tokimasa
Tags: ellipses, precalc
Tokimasa
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#1
May3-05, 02:59 PM
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My math teacher has been out for 2 days and the subs have just been putting notes up and I'm totally not getting this. So maybe someone here can explain everything in other ways. Note that my book does a crappy job of explaining this subject. But it does break it down into two distince areas - Ellipses Centered at (0,0) and Ellipses Centered at (h,k).

Ellipses Centered at (0,0)

According to my book, there are 6 points that make up an ellipses. V and V' are the vertices of the ellipse where V has the coordinates (a,0) and V' has the coordinates (-a,0). The points m and m' represent the points on the minor axis of the array and have the coordinates (0,b) and (0,-b) respectivally. The ellipse also has two foci (F and F') located at (c,0) and (-c,0). The value "a" represents the distance from the center of the array to the vertex on the major axis. The value "b" is the distance from the center to the vertex on the minor axis. The value "c" represents the distance from the center to the focus.

The book also goes on to give the equation of an ellipse is (x^2/a^2) + (y^2/b^2) = 1 and gives the formula a^2 = b^2 + c^2 for finding the value of c (which turns into c^2 = abs(a^2 - b^2)).

The book goes on to say that you must set up the formulas such that a^2 > b^2. But that doesn't make any sense. I try to plug everything in for an ellipse that I know has a major axis on the y axis (meaning it is vertical) yet when I graph it, it looks like it's horizontal (its major axis was on the x axis).

So can someone explain a universal method for solving problems involving ellipses centered at (0,0)?

Ellipses Centered at (h,k)


Most of the questions that I had for ellipses centered at (0,0) apply here, but I have some more questions.

Exactly what do these steps mean:
(1) collext x terms/y terms on one side
(2) get constant on the other side
(3) complete the square
(4) factor
(5) divide to get a constant of 1
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Pyrrhus
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#2
May3-05, 03:33 PM
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An elipse is a geometrical place (It means that constituents points of the curve satisfy a condition or more) of a point that moves in the plane in such a way that the sum of its distance to two fixed points (focal points) is always constant and bigger than the distance between both fixed points. Such distance will be equal to the major axis of the ellipse.

This means that the distance between a F' focal point to a point (x,y) + the distance from the other F focal point to the same point (x,y) is constant (2a = Major axis).

From that you can derive the Ellipse Equation, or relation F(x,y).

The proof is at http://mathworld.wolfram.com/Ellipse.html

An ellipse at (h,k) it's just a Ellipse at (0,0) that has geometrical transformation, a traslational one. So you can simply add the focus, major axis, minor axis, center, and then do the traslational transformation by adding the coordinates (h,k) to each of the points.
Integral
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#3
May3-05, 03:39 PM
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It is not clear what your trouble with the first part is, seems to me you have a pretty good handle on it. Perhaps you could post your work, we may be able to give more specific help.

As for the second question. In general the equation for an ellipse not centered on the origin will have the form

[tex] ax^2 + bx + c + dy^2 +ey + f =0 [/tex]

As given in your problems it may not be ordered as above. So your first task is to rearange terms to get the above form.

note that this is simply the general form of a quadradic equations in x and y, Next you will need to express it in the general form of an elipse:

[tex] \frac {(x-h)^2} {a^2} + \frac {(y-k)^2} {b^2} = 1 [/tex]

To get from the first to second form you may need to complete the squares in each of the x and y equations. When that is done you will have something like this

[tex] C_1 (x-h)^2 + C_2 + C_3(y-k)^2 + C_4 =0[/tex]

Where the [itex] C_n[/itex] are different constants. Now move the constants I have called [itex] C_2 [/itex] and [itex]C_4[/itex] to the right hand side, then divide both sides by [itex] C_2 + C_4[/itex] to get the standard form.

Tokimasa
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#4
May3-05, 03:47 PM
P: 17

PreCalc - Ellipses


Other than graphing, how do you determine the orientation of the major axis of an ellipse? I've been solving for y and using my TI calc.

How do you solve for c?

When you have a, b, and c - How do you determine exactly where the points lie? (a,0) won't work for a vertical ellipse. So do you just flip it to (0,a)? Or is there a method that works? My problem is finding the specific points that make up the ellipse (F, V, m).
Pyrrhus
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#5
May3-05, 03:52 PM
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The orientation fo ellipse can be obtained by looking at the equation in the standard form

[tex] \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 [/tex]

Where a is, gives you the orientation, in my example, The ellipse will have a horizontal major axis.

Can you post an example you are having problem to locate the points?

It will make it easier.
Pyrrhus
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#6
May3-05, 04:20 PM
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My advice for Ellipses centered at (h,k) is to solve it as it was centered at (0,0) and then do a traslational transformation to each of the points of the elements of the elipse (Major Axis, Minor Axis, Focal Points) to the coordinates (h,k).
Tokimasa
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#7
May3-05, 04:31 PM
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So what formula should I use to find c?

My choices are (c^2 = abs(a^2 - b^2)) or the original formula that requires a^2 > b^2 which is (a^2 = b^2 + c^2).
Pyrrhus
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#8
May3-05, 04:39 PM
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Use [itex] c^{2} = a^2 - b^2 [/itex] with the condition [itex] a>b [/itex]
Tokimasa
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#9
May3-05, 05:46 PM
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You people have been helpful. But I didn't see an answer to one of my questions.

When you have a, b, and c - How do you determine exactly where the points lie? (a,0) won't work for a vertical ellipse. So do you just flip it to (0,a)? Or is there a method that works? My problem is finding the specific points that make up the ellipse (F, V, m).
Pyrrhus
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#10
May3-05, 06:03 PM
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If it's centered at the origin, i don't see a problem finding the points, they are basicly the distances, because the curve cuts the x and y axis.
Pyrrhus
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#11
May3-05, 06:14 PM
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For Example let's suppose

[tex] \frac{(x-9)^2}{36} + \frac{(y-6)^2}{9} = 1[/tex]

We do it as it's centered at the origin so

[tex] a = 6 [/tex]
[tex] b = 3 [/tex]
[tex] c = \sqrt{27} = 3 \sqrt{3}[/tex]

The ellipse is horizontal therefore

For the major axis the points are

(6,0) and (-6,0)

For the minor axis the points are

(0,3) and (0,-3)

For the focal points

([itex] 3 \sqrt{3} [/itex],0) and ([itex] -3 \sqrt{3} [/itex],0)

Now we do a tralastion transformation

we know the actual center for the ellipse is (9,6)

So each of the points of its elements are actually located at

For the major axis the points are

(6+9,6) and (-6+9,6)

For the minor axis the points are

(9,3+6) and (9,-3+6)

For the focal points

(9 + [itex] 3 \sqrt{3} [/itex],6) and (9 + [itex] -3 \sqrt{3} [/itex],6)


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