# Help graphing the equation of an ellipse.

1. Jul 8, 2006

### alancj

I’m having trouble graphing this equation:

16x^2+9y^2=144

I have entered it into both my Algebrator program and the quickmath.com website. They both gave the same graph of the ellipse. In each case the major axis was vertical. With the Algebrator program it simplifies the above equation into (quickmath.com didn’t do this):

(x^2)/9 + (y^2)/16 =1

This makes sense; just divide each side by 144 to make it all equal to 1. However, my textbook says that a^2 is always >/= to b^2. So the simplified equation seams to be wrong in the first place. If I try and use it to graph an ellipse myself then I find that the major axis would actually be shorter than the minor axis!

According to my textbook when the x term comes first the direction of the major axis is horizontal not vertical as was graphed by both programs.

In either case (horizontal or vertical form of the equation) the first term’s denominator is supposed to be a^2, and the next term’s denominator is b^2. That would mean that a=3 and b=4 and the length of the major axis would be 2a (6) and the minor would be 2b (8).

So this doesn’t make any sense, am I supposed to just graph this and ignore the fact that 2a is hardly “major” and 2b is hardly “minor”?

I don’t know of any other way of doing this problem. Any help would be much appreciated.

Thanks,
Alan

2. Jul 8, 2006

### arildno

You have:
$$(\frac{x}{3})^{2}+(\frac{y}{4})^{2}=1$$
It should be very easy for you to find out what is major and minor axes in this form.

3. Jul 8, 2006

### Beam me down

$a^2$ does NOT have to be $\geq b^2$. The major and minor axis are only depedndant on the values of a and b, not the order the terms appear. As such:

$$(\frac{x}{3})^{2}+(\frac{y}{4})^{2}=1 \Leftrightarrow (\frac{y}{4})^{2}+(\frac{x}{3})^{2}=1$$

Where $\Leftrightarrow$ means equivelent or "the same as". So the implicit function for the ellipse is independant of the order of the x and y's. This is the basic law of addition where x+y is the same as y+x.

Last edited: Jul 8, 2006
4. Jul 8, 2006

### alancj

I already know what the major axis is. Did you read my post? 2a is supposed to be the major axis, 2b is supposed to be the minor. a^2 is below the first (x) term and b^2 is below the second (y) term. The major axis is 2a=6 and the minor axis is 2b=8. Can you tell me why the MAJOR axis is SMALLER than MINOR axis? Isn't it supposed to be the other way around? This would result in an ellipse with its REAL major axis being vertical instead of horizontal. My equation’s first term starts with an x. The standard form of the equation whose x term is first is supposed to have the direction of the major axis horizontal, not vertical. Therein lies the dilemma.

-Alan

5. Jul 8, 2006

### arildno

No. You have completely misunderstood this.
The major axis is the greatest one; that is not dependent upon whether the associated variable is called "x" or "y".
In this case, the major axis is, indeed, along the vertical.

6. Jul 8, 2006

### alancj

According to my textbook the major axis is 2a. The minor is 2b. a^2 is 9, is it not? b^2 is 16, is it not? Then it would be reasonable to conclude that 2a=6 and 2b=8, which doesn't make much sense.
And according to my textbook the form of the equation x^2/a^2 + y^2/b^2 = 1 has the major axis horizontal. My equation is in that form.

Last edited: Jul 8, 2006
7. Jul 8, 2006

8. Jul 8, 2006

### HallsofIvy

Read your textbook again! The "major" axis is always the longer of the two axes, the minor axis is the shorter (That's why they call them "major" and "minor"! ) whether that longer axis is vertical or horizontal. However, the longer axis may be eithervertical or horizontal (or even at an angle but you probably won't see those in this course). If the major axis is horizontal then the "a" in $\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$ will be larger than the "b". If the major axis is vertical then b will be larger than a.

By the way, since the title of this was "Help graphing the equation of an ellipse", here's a quick way to do it by hand: with the ellipse equation in "standard form",
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$$
draw vertical lines through (a,0) and (-a,0) and horizontal lines through (0,b) and (0,-b). Now sketch your ellipse so that it is tangent to those lines at (a,0), (-a,0), (0,b), and (0,-b).

Last edited by a moderator: Jul 8, 2006
9. Jul 8, 2006

### alancj

Yes it does! I've had it in front of me the whole time. :rofl:

10. Jul 8, 2006

### Beam me down

Please check the http://en.wikipedia.org/wiki/Ellipse" [Broken]. Of importence in this case is:

In less technical terms the major axis is the one which passes through the foci of the ellipse (a simple definition is the points where the most curved section of the ellipse is centred on, there are two in a normal ellipse). The line passing through the two foci and stopping on the ellipse is always the longest straight line that can be drawn in the ellipse, so it is called the major axis.

Last edited by a moderator: May 2, 2017
11. Jul 8, 2006

### arildno

Note that your textbook uses the convention that "a" is ALWAYS the largest number, "b" the least.

In many computer applications, however (and most math books), "a" is used for the denominator in the x-fraction, whether or not "a" is the largest number.

Either of these conventions is, by its own, perfectly acceptable, but you must not mix them together!!

Furthermore, pay particular attention to what your textbook calls "standard form".

Given:
$$\frac{x^{2}}{3^{2}}+\frac{y^{2}}{4^{2}}=1$$
This is NOT standard form according to your textbook, but is standard form for most other books!

Standard form for your textbook is gained by interchanging the x-and y-terms:
$$\frac{y^{2}}{4^{2}}+\frac{x^{2}}{3^{2}}=1$$

Last edited: Jul 8, 2006
12. Jul 8, 2006

### alancj

why did you switch the two? I'm confused.

13. Jul 8, 2006

### arildno

1. The sum of two numbers is the same, irrespective of the order of those numbers (3+4=4+3, 4.555557+3.01=3.01+4.555557 and so on)
Thus, we have the identity (not equation!):
$$\frac{x^{2}}{9}+\frac{y^{2}}{16}=\frac{y^{2}}{16}+\frac{x^{2}}{9}$$
2. Your book requires that in STANDARD FORM, the number with the largest denominator is to be written first in the equation.

14. Jul 8, 2006

### alancj

sooo... I don't determine what is a^2 and what is b^2 by what term they appear under? I just look at the denominator's and call the bigger one a^2?

WTF! Why couldn't the book just have said that! I don't know how I would have seen that if I didn't ask. I think that I hate my textbook...

Now that I've re-read the replies I think that's what you guys have been saying all along. Here I thought you guys were misunderstanding me, and it was actually the other way around!

So, the graph is right; my equation is just the same as saying y^2/16 + x^2/9=1, and a^2 is actually bigger than b^2 like it is supposed to be! I totally didn’t see that until arildno’s last post with the identity (not) equation side by side.

Thanks,
Alan

15. Jul 8, 2006

### arildno

Correct!

It does, in a convoluted implicit way (it states that 2a is the MAJOR axis; from this, it is deducible that "a" is the name of the biggest number)[/QUOTE]
It certainly makes up some rather odd conventions.
Read your book CAREFULLY. Try to think over: Have I really understood this?

Glad I was of some assistance!

16. Jul 9, 2006

### HallsofIvy

Your book DID say that! Under "key concepts" it has two columns: one has
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}= 1$$
with "major axis horizontal" under it. The other column has
$$\frac{y^2}{a^2}+ \frac{x^2}{b^2}= 1$$
with "major axis vertical" under it.

Don't you hate when that happens!

Also note that many, perhaps most, texts always have the form
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$$
allowing either a or b to be larger.