Ellipses

- Babynini

Ellipses

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- Babynini

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- Babynini

@freckles

- Babynini

@rational

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## More answers

- Babynini

oh I know where I calculated wrong.
in that
c/a and forgot that a was =4 and not 2

- Babynini

would it be
b^2 = 2?

- Babynini

nope that's not right either.

- Babynini

@rvc do you know how to do these?

- Babynini

@jim_thompson5910

- rvc

bad at circle and conics
sorry :(

- Babynini

Thanks for looking at it though. Ah well.

- rvc

@divu.mkr help please :)

- Babynini

first off we find a from the major axis
a= 4
then we find c by looking at the eccentricity
c/a = sq7/2

- anonymous

focii on y axis means that the major axis is on Y aix and you are given with the length of major axis and ecc.. you can find minor axis from that data

- anonymous

axis**

- alekos

It would have a vertical major axis so (x-h)^2/b^2 + (y-k)^2/a^2 =1
where 2b=8

- Babynini

how do I find the minor axis in there?

- alekos

eccentricity = sqrt(a^2-b^2)/a

- anonymous

b^2 = a^2(1-e^2)

- alekos

so from this you can find the minor axis

- Babynini

hmm

- alekos

focii are on the y-axis so h=0

- Babynini

so
x^2/8 + y^2/16 = 1

- Babynini

is that the answer?

- Babynini

@iambatman save me. lol

- anonymous

Hollllllaaaa

- Babynini

aloo

- anonymous

major axis = 8 ye?

- Babynini

I just need to find an equation :P

- Babynini

yeah

- Babynini

major axis = 2a
so a = 4

- anonymous

Yes good :P
\[e = \frac{ c }{ a } ~~~\text{and}~~~ c^2 = a^2-b^2\]
yes yes this looks good, it will work out nicely

- anonymous

Ok so notice we're given the eccentricity (e) and it tells us the Foci is on the y - axis, this is important because the foci are ALWAYS on the major axis. So will the major axis run vertically or horizontally?

- Babynini

vertically?

- anonymous

That sounds good, so what should we do next

- anonymous

You should already realize an error in your work with that information

- Babynini

a should be the bigger one?

- anonymous

You're too good :)

- Babynini

Aw, shucks.

- Babynini

so what numbers do i flip around?

- anonymous

Well lets find b first

- anonymous

To do that we have \[e = \frac{ c }{ a }\] and \[c^2 = a^2 - b^2\]

- Babynini

e = sq7/2
but we're given that a = 4 so do we double that whole thing?

- anonymous

What was that cute game we were playing earlier? I had a name for it

- Babynini

lol substitution?

- anonymous

Yeah that's it ;)

- Babynini

but we're given the e!

- anonymous

So what's stopping you from solving for c :)

- Babynini

no b?

- anonymous

We have a and e!

- Babynini

ooh oh

- Babynini

sq7/2=c/4

- anonymous

:))))

- anonymous

So what will be our c value?

- Babynini

sq14?

- Babynini

\[2\sqrt{7}\]

- anonymous

Looks good, now what's our next step :)

- Babynini

find b!

- anonymous

Right, and how will we do that

- Babynini

so that's just c, not c^2

- anonymous

Yes good observation!

- Babynini

so now
c^2=a^2+b^2

- anonymous

Stop right there for a second!

- Babynini

o.o yes sir.

- anonymous

Lets look at the equation you said, what is that for :)

- Babynini

c

- anonymous

ircle

- Babynini

lol

- anonymous

That's where a lot of people make mistakes, for an ellipse it's \[c^2 = a^2 - b^2\]

- anonymous

Note the negative sign

- Babynini

o.o gasp. that is no bueno.

- Babynini

b^2=a^2-c^2 then

- anonymous

:), now remember as I always say it's a game of substitution

- Babynini

\[b^2 = 16 - (2\sqrt{7})^2\]

- Babynini

\[b^2 = 16-28\]
\[b^2=-12\]

- Babynini

\[b=\sqrt{12}\]

- anonymous

No, the negative is there, something is wrong

- anonymous

Err, \[e = \frac{ c }{ a } \implies c = e \times a \implies c = \frac{ \sqrt{7} }{ 4 } \times 4 \implies \sqrt{7}\] bleh :)

- Babynini

..oh hehe

- anonymous

I guess I should be doing the math as well haha

- Babynini

sorry!

- Babynini

ai ai ai I need to go to bed soon. waay to early on a school night =.= and you do too.

- anonymous

No it's alright we just had mistaken the eccentricity we had sqrt(7)/2 when it's 4 :P

- Babynini

oh xD

- anonymous

Don't worry about me :), I'm the dark knight, night is where I work muahah!

- Babynini

right, right, how could I forget!

- anonymous

Alright so lets find b again bleh :)

- anonymous

\[b^2 = a^2-c^2\]

- anonymous

\[a = 4~~~c = \sqrt{7}\]

- Babynini

b = 3

- Babynini

b^2 = 9

- anonymous

Now that looks good

- Babynini

so..the answer I had at the very beginning. Look at the screenshot thing =.=

- anonymous

major axis runs vertically :P

- Babynini

ooh myy goooshh aspidghpaihgpi

- Babynini

fricken hour later hahahah

- anonymous

It's ok, the struggle is more important than the final answer, you will remember how to do it now :)

- Babynini

we're geniuses! :)

- Babynini

for sure though!

- anonymous

Not me, you are a genius! You deserve a fields medal hehe!

- Babynini

noo no bro the answers wrong still
D:

- anonymous

What did you put

- Babynini

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- anonymous

Nope.

- anonymous

a = 4, a^2 = 16

- Babynini

...can you write out the whole answer just so i make sure to put it right?

- anonymous

and y runs vertically hence \[\frac{ x^2 }{ 9 }+ \frac{ y^2 }{ 16 } = 1\]

- Babynini

That little green mark though :')

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- anonymous

Yes, sorry about that I did not check if you put a^2

- Babynini

it's all good :) so it was indeed my first answer just forgot to ^2 the a

- anonymous

Yes, we just had to switch it around, but in a test you would have to show your work, and you now know how to do that! :)
It's a game of substitution as always!

- Babynini

haha cuz we hate the players (not really) but love the game ;)

- Babynini

yes indeed!

- Babynini

*hug hug hug. you are a trooper. haha and with that. I am off to bed.

- anonymous

Take care :)

- Babynini

Night, Batman.

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