Question

What conic section is this?

Answer 1

32x^2 + 9y^2 = 324

Answer 2

whenever you have x^2 and y^2 and everything else is degree 2 or less, it's a conic section.
If the coefficients of x^2 and y^2 are both positive and identical, it's a circle.
If the coefficients of x^2 and y^2 are both positive and different, it's an ellipse.
If the coefficients of x^2 and y^2 are different in sign, it is a hyperbola, which \(could\) degenerate into two straight lines.

Answer 3

What about a parabola?

Is this one an ellipse?

Answer 4

yes, this one is an ellipse

a parabola has an x^2 and a y (or vice versa)

Answer 5

Okay... I'm supposed to write it in standard form now. How would I do that?

Answer 6

I'm still having trouble with this problem... does anyone know what to do?

Answer 7

For an ellipse?
Standard form is with a 1 on the right side, ya? :)

So divide both sides by 324.

Answer 8

Zeppy coming to the rescue!

32/324x^2 + 9y^2/324 = 1

Answer 9

mm k great!
See if 32 and 324 have anything in common.
And do likewise with the 9 and 324.
We'd like this to be in the form x^2/a^2, number only in the bottom.

Answer 10

4/81x^2 + 3/108y^2 = 1

Can this be simplified any further?

Answer 11

Oh right, so the x^2 is in the numerator

Answer 12

Same with the y^2

Answer 13

That's as far as they broke down?
Hmm that's annoying :d

We'll have to do something fancy with our fractions,
you're not going to like this.

Answer 14

Haha :D

Actually, I know what the answer should be, I just don't know how to get it there. So let's do this!

Answer 15

\[\large\rm \frac{4x^2}{81}+\frac{9y^2}{324}=1\]Let's leave the y stuff un-simplified.
If this isn't clear why just yet, that's ok.

Maybe you notice something special about all of these numbers?

Answer 16

Hmmm. Nothing jumps out at me...

Answer 17

Wait! They are all perfect squares

Answer 18

Mmm good good good.
And remember our ellipse formula wants us to write the denominators as perfect squares!! ya? x^2/(a)^2

Answer 19

2x^2/9 + 3y^2/18 = 1

Is that right?

Answer 20

\[\large\rm \frac{4x^2}{81}+\frac{9y^2}{324}=1\]
I'm going to apply a nasty fraction rule that you're going to hate.
So here it goes,
\[\large\rm \frac{x^2}{\left(\frac{81}{4}\right)}+\frac{y^2}{\left(\frac{324}{9}\right)}=1\]

Answer 21

What do you think?
Did your brain explode? :(

Answer 22

I can't think of any more quality Abble puns.. ugh

Answer 23

You can write it as 2^2x^2/9^2+3^2y^2/18^2=1
writing all the squares out like that.

But I was going to save that for the very last step,
otherwise I think it's a little confusing.

Answer 24

Or as (2x/9)^2+(3y/18)^2=1

Which maybe is ok to work with.

Answer 25

Oh oh oh we can do the division on the y's,
we just didn't go far enough, my bad.
3 goes evenly into 108, giving us 36 in the denominator.

Answer 26

Bahhh ok we'll worry about that in a sec.
I wanna know what you think of that fraction shinanigans :P

Answer 27

Hahaha okay so what's our next step? I have a feeling this is about to get hairy.

Answer 28

Then we can write our denominators as perfect squares,\[\large\rm \frac{x^2}{\left(\frac{9}{2}\right)^2}+\frac{y^2}{\left(\frac{18}{3}\right)^2}=1\]

Answer 29

So all you did was move the coefficient on the numerator below the coefficient on the denominator?

Answer 30

Mmmm uh oh :(
I should've checked your arithmetic earlier :(
I think we have the wrong values.

Answer 31

Oh no!

Answer 32

\[\large\rm \frac{32x^2}{324}+\frac{9y^2}{324}=1\]I think we're left with ... an 8 on the first fraction, ya?
Not the 4.

Answer 33

Face palm.

Yep, I think you're right...

Answer 34

\[\large\rm \frac{8x^2}{81}+\frac{y^2}{36}=1\]

Answer 35

Oh boy, that makes things more difficult :\ hmm

Answer 36

Same procedure though?

Answer 37

Yes.
We'd like to get the `number` completely under the x.

Answer 38

\[x^2/(9^2/\sqrt8) + y^2/6 = 1\]

Something like this?

Answer 39

Here is how I did that process, just in case it is not clear,\[\large\rm \frac{8x^2}{81}\quad=x^2\cdot\frac{8}{81}\]From here, I'm basically applying my "keep change flip" rule in reverse,\[\large\rm =x^2\div \frac{81}{8}\]Which I'm writing this way,\[\large\rm \frac{x^2}{\left(\frac{81}{8}\right)}\]

Answer 40

Yes, very good :)
The 6 is being squared also.

Answer 41

Right! Good good :)

Is that it?

Answer 42

Oh the first fraction also,
let's clean that up a little.
It's not just the 9 being squared but also the sqrt8.

So your final result should look something like,\[\large\rm \frac{x^2}{\left(\frac{9}{\sqrt8}\right)^2}+\frac{y^2}{6^2}=1\]Which is in the standard form of,\[\large\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]And so we can clearly see that our a=9/sqrt8 and b=6.

Answer 43

How would I graph something like that?

Answer 44

It's centered at the origin... and it's a horizontal ellipse, right?

Answer 45

Looks like it's going to be vertical since 6 > 9/sqrt8

Answer 46

I thought a^2 was always the major axis?

Answer 47

Is that how it usually goes?
Ok then I guess we would call a=6 and b=9/sqrt8, my bad

Answer 48

So it would be a vertical ellipse centered at the origin, 3.18 units horizontally on each side and 6 units up from the origin?

Answer 49

Oh okay

Answer 50

Cool, I think I got it :)

This question actually had a bunch of other steps, so I'll open up a new Q. It gets even more complicated :O

Answer 51

not too hard to graph from that point, ya?
Getting it into standard form can be some tough work though D: