Question

Find the equation of the ellipse with vertices (0, -1) and (12, -1) and minor axis of length 6

Answer 1

So, I'd start with the standard form of an ellipse. Once you have that, you just have to know what you're looking for and insert what you find into the right spot, really.
Graphing what you know also helps to find stuff like the \(a\) and \(b\) from the major and minor axes.

Answer 2

TF1 + TF2 = K?

Answer 3

Hmm.. have you learned \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]
That formula? At least, I am more familiar with this equation. :P

Answer 4

no haha but i will use it.. so which x and y point do i plug in?

Answer 5

In that equation form, we are looking for (h, k), the center of the ellipse, "a", the distance from the center to the vertex, and "b", the distance from the center to a covertex.

Of course, the center is really easy -- it's the midpoint of the two vertices
The a-value is that distance from the center to the vertex, so that's fairly straightforward.
We can then use 1/2 of the minor axis for the b-value, and its just fill-it-in from there

If you're familiar with using the other formula, I'll have to go search up some resources on using that one. lol

Answer 6

no its okay! i like this one better

Answer 7

but i still don't understand what plugs into the x and y

Answer 8

We don't need to plug anything in for x and y. This equation will look like this:when we find all the information:
\[ \frac{(x - \#)^2}{\#} + \frac{(y - \#)^2}{\#} = 1 \]
Where the #'s are just numbers.

Answer 9

so we only know b = 3

Answer 10

Yep. We can also find the center with the two vertices because the center is just the midpoint.
(0,-1) to (12,-1), (6,-1) is the midpoint (just by taking 1/2 of (12 + 0), pretty much counting)
That is the center, so it goes into the equation like (h, k) = (6, -1)

Answer 11

so [(x-6)^{2}\div a ^{2} \times(y--1)^{2}\div 3 ?\]

Answer 12

\[[(x-6)^{2}\div a ^{2} \times(y--1)^{2}\div 3] ?\]

Answer 13

\[ \frac{(x-6)^2}{a^2} + \frac{(y + 1)^2}{3^2} = 1 \]
It's actually an addition between them, and we just have to square that 3 in the bottom because the original formula is "b^2", b=3

Answer 14

gotcha

Answer 15

Then, a is actually the distance from the center to the vertices, which we pretty much counted to be 6 units when we were finding the center. (0,-1) to (6,-1) is 6
\[ \frac{(x - 6)^2}{6^2} + \frac{(y+1)^2}{3^2} = 1 \]
And that's it. We can clean up the exponents in the denominator and it should be good.
\[ \frac{(x - 6)^2}{36} + \frac{(y + 1)^2}{9} = 1 \]

Answer 16

And thats it??

Answer 17

Yeah, that's the equation for the ellipse using the formula I had.

Your standard form is a little different from mine though. From what I gather, its the sum of the distances from some point to each foci is equal to the major axis' length.

Answer 18

Thats way easier than what i've been doing. Thank you!

Answer 19

Yeah. I did this section rather recently and we skipped over the formula that you posted. ;)
The only thing with my formula is that you have to put a^2 under x or y depending on which is the major axis (contains the vertices) and the minor axis.
I used the one for this scenario, with the line parallel to x-axis as the major axis

Answer 20

"Horizontal major axis" was the term I was searching for! We had a horizontal major axis, so the x-value had the a^2 under it. If the major axis was vertical, then we have it under the y.

Answer 21

\[ \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1\]
Major vertical axis case

Answer 22

This is why I stink at math.. too many equations to remember

Answer 23

Yeah, memorizing the equations can be a chore.

Just curious, for the method you use, did you have to find the focus points to get the equation?

Answer 24

well like for x^2/9 + y^2/16 = 1 yeah.. some of them you could just look at the graph

Answer 25

I'd find foci using this property and Pythagorean theorem, to actually figure out the equation using the formula "TF1 + TF2 = 2A" from the points given. D:
We can find center the same, find a and b the same, except then we have to find c, get F(h + c, k) and F(h - c, k) (horizontal major axis) or F(h, k + c) and F(h, k - c) (Vertical major axis).

Then take distance from (x,y) to F1 and distance from (x,y) to F2 and set equal to 2a.... or so my intuition would say from reading some other sites. That seems like a lot of work to me. :P