Ellipses

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oh I know where I calculated wrong. in that c/a and forgot that a was =4 and not 2
would it be b^2 = 2?
nope that's not right either.
@rvc do you know how to do these?
  • rvc
bad at circle and conics sorry :(
Thanks for looking at it though. Ah well.
  • rvc
@divu.mkr help please :)
first off we find a from the major axis a= 4 then we find c by looking at the eccentricity c/a = sq7/2
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
axis**
It would have a vertical major axis so (x-h)^2/b^2 + (y-k)^2/a^2 =1 where 2b=8
how do I find the minor axis in there?
eccentricity = sqrt(a^2-b^2)/a
b^2 = a^2(1-e^2)
so from this you can find the minor axis
hmm
focii are on the y-axis so h=0
so x^2/8 + y^2/16 = 1
is that the answer?
@iambatman save me. lol
Hollllllaaaa
aloo
major axis = 8 ye?
I just need to find an equation :P
yeah
major axis = 2a so a = 4
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
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?
vertically?
That sounds good, so what should we do next
You should already realize an error in your work with that information
a should be the bigger one?
You're too good :)
Aw, shucks.
so what numbers do i flip around?
Well lets find b first
To do that we have \[e = \frac{ c }{ a }\] and \[c^2 = a^2 - b^2\]
e = sq7/2 but we're given that a = 4 so do we double that whole thing?
What was that cute game we were playing earlier? I had a name for it
lol substitution?
Yeah that's it ;)
but we're given the e!
So what's stopping you from solving for c :)
no b?
We have a and e!
ooh oh
sq7/2=c/4
:))))
So what will be our c value?
sq14?
\[2\sqrt{7}\]
Looks good, now what's our next step :)
find b!
Right, and how will we do that
so that's just c, not c^2
Yes good observation!
so now c^2=a^2+b^2
Stop right there for a second!
o.o yes sir.
Lets look at the equation you said, what is that for :)
c
ircle
lol
That's where a lot of people make mistakes, for an ellipse it's \[c^2 = a^2 - b^2\]
Note the negative sign
o.o gasp. that is no bueno.
b^2=a^2-c^2 then
:), now remember as I always say it's a game of substitution
\[b^2 = 16 - (2\sqrt{7})^2\]
\[b^2 = 16-28\] \[b^2=-12\]
\[b=\sqrt{12}\]
No, the negative is there, something is wrong
Err, \[e = \frac{ c }{ a } \implies c = e \times a \implies c = \frac{ \sqrt{7} }{ 4 } \times 4 \implies \sqrt{7}\] bleh :)
..oh hehe
I guess I should be doing the math as well haha
sorry!
ai ai ai I need to go to bed soon. waay to early on a school night =.= and you do too.
No it's alright we just had mistaken the eccentricity we had sqrt(7)/2 when it's 4 :P
oh xD
Don't worry about me :), I'm the dark knight, night is where I work muahah!
right, right, how could I forget!
Alright so lets find b again bleh :)
\[b^2 = a^2-c^2\]
\[a = 4~~~c = \sqrt{7}\]
b = 3
b^2 = 9
Now that looks good
so..the answer I had at the very beginning. Look at the screenshot thing =.=
major axis runs vertically :P
ooh myy goooshh aspidghpaihgpi
fricken hour later hahahah
It's ok, the struggle is more important than the final answer, you will remember how to do it now :)
we're geniuses! :)
for sure though!
Not me, you are a genius! You deserve a fields medal hehe!
noo no bro the answers wrong still D:
What did you put
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Nope.
a = 4, a^2 = 16
...can you write out the whole answer just so i make sure to put it right?
and y runs vertically hence \[\frac{ x^2 }{ 9 }+ \frac{ y^2 }{ 16 } = 1\]
That little green mark though :')
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Yes, sorry about that I did not check if you put a^2
it's all good :) so it was indeed my first answer just forgot to ^2 the a
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!
haha cuz we hate the players (not really) but love the game ;)
yes indeed!
*hug hug hug. you are a trooper. haha and with that. I am off to bed.
Take care :)
Night, Batman.

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