How do you know when an ellipse is parallel to the x-axis or y-axis? For example, If I'm given 4x^2-16x+9y^2+18y-11=0 and I simplify that into an equation of an ellipse: [(x-2)^2]/[9] + [(y-1)^2]/[4] = 1, how do I know if the ellipse is parallel to the x-axis or parallel to the y-axis?

Question

jtvatsim · Answer

Not exactly sure what you mean by "parallel." Maybe you mean whether the ellipse is wide along the x-axis or tall along the y-axis? :)

jtvatsim · Answer

As in this vs. this?|dw:1439526085570:dw|

anonymous · Answer

Well, Im trying to solve this problem: 

Find the center, vertices, and foci of the ellipse given by
4x^2 - 16x + 9y^2 + 18y - 11 = 0

I have figured everything out till:

 [(x-2)^2]/[9] + [(y-1)^2]/[4] = 1

Now i have to find the foci and vertices, but there are two different expressions used to find the foci and vertices, and they both depend on whether your ellipse is parallel to the x-axis or y-axis...

jtvatsim · Answer

OK, I get what you mean. I had never heard that terminology before, but we are both thinking of the same thing. I believe you should have (y+1)^2/4 in your expression rather than (y-1)^2/4, but aside from that... whichever expression has the larger denominator will be the major axis. Since your expression for x^2 has the larger denominator 9, the ellipse will be parallel to the x-axis.

jtvatsim · Answer

In fact, the denominators are the radius of the axis squared. So in this case, the major axis has a radius of 3 along the x-axis, since 3^2 = 9. The minor axis has a radius of 2 along the y-axis, since 2^2 = 4.

anonymous · Answer

|dw:1439526368806:dw| This is what I'm looking at..

jtvatsim · Answer

Sorry, openstudy didn't tell me that you had replied.

jtvatsim · Answer

OK, yes my explanation above still stands. It is the "major axis" that is parallel, not the "ellipse" that is parallel just for future terminology. Do you understand the table though?

jtvatsim · Answer

Remember that the general equation of an ellipse is represented by
$$\frac{(x-h)^2}{first\ denominator^2} + \frac{(y-k)^2}{second \ denominator^2} = 1$$
and we make "a" the bigger of the two denominators, while "b" is the smaller of the two denominators according to your table.

So, in your case, we have
$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$
since the x-denominator was bigger.

jtvatsim · Answer

What you should have for your question is that
a = 3
and
b = 2

anonymous · Answer

Thank you very much, I worked the rest out and it matched the answer key. This is valuable information, I appreciate it! @jtvatsim

jtvatsim · Answer

Great job! I'm glad you figured it out! :)