help me plz

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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brb in 20

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im back
first of all you should know is it horizontal or vertical a always greater than b `a>b` if a is under the x variable then it would be `horizontal equation` \[\rm \frac{ (x-h)^2 }{\color{ReD}{ a^2} }+ \frac{ (y-k)^2 }{ b^2 }=1\] if a is under the y variable then it would be `vertical equation` \[\rm \frac{ (x-h)^2 }{ b^2 }+ \frac{ (y-k)^2 }{\color{reD}{ a^2} }=1\]
so is it vertical or horizontal ?
i think its vertical
foci for horizontal \[\large\rm (h, k \pm c)\] and for vertical equation \[\large\rm (h \pm c , k)\] use the equation to find c \[\huge\rm c^2 =a^2-b^2\]
ok so which one do i use
well a is bigger number which is under the x variable so is it vertical or horizontal ?
read my first comment :=)
so it is horizontal right
yes right \[\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1\] now use the equation posted above to find c
c^2=a^2-b^2 this one
wait what
foci for horizontal \[\large\rm (h, k \pm c)\] and for vertical equation \[\large\rm (h \pm c , k)\] those two are foci pair we need so use this equation to find c \[\huge\rm c^2 =a^2-b^2\]
a^2 and b^2 are in the given equation
wow ok so umm can you start me off plz
`horizontal equation` for ellipse \[\rm \frac{ (x-h)^2 }{\color{ReD}{ a^2} }+ \frac{ (y-k)^2 }{ b^2 }=1\] the number at the denominator are a^2 and b^2 and like i said bigger number equal to a^2
what two numbers are replaced with a^2 and b^2 in this equation \[\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1\] ??
brb i gtg cook dinner
36 and 11
right which one is a^2 and which one is b^2 ?
a^2 is 36 b^2 is 11
{ (x-h)^2 }{36} }+/{ (y-k)^2 }{11}=1 is this the correct formula
this ellipse is centred on the origin yep?
yep
|dw:1444085734817:dw|
yep
Para-phrasing Wiki: The distance from the origin O to either focus is f where: \[f = \sqrt{a^2-b^2}.\]
ok she gave me a diffrent formula soo what do i do im so confused now
here, what is a and what is b? general equation for ellipse centred on Origin is: \(\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\)
a = ?? b = ??
then |dw:1444086259207:dw|
(x-h)^2 }{36} }+/{ (y-k)^2 }{11}= ok so i got this
so that means you have finished it?!
no i have not thats the first equation
(y-k)^2 {11} this is my answer
OK for an equation centred on O, we have general equation: \[\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\] The focii are at a distance f from O, where \[f = \sqrt{a^2-b^2}\] |dw:1444086808815:dw|
soz, we crossed posts. i understood the question to be finding the focii of the ellipse
yes idk what im doing im so cunfused here im sorrry
no worries you have here: \(\large\rm \frac{ x^2 }{ \color{ReD}{6^2} } +\frac{ y^2 }{(\sqrt{\color{red} 1\color{red}1 })^2}=1\) and generally speaking: \(\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\) and: \(\large f = \sqrt{a^2-b^2}\)
so f =????
omg i have no clue on how to do this like wth
does f equal -4
\(a = 6, \; b = \sqrt{11}\) \(f = \sqrt{ 6^2 - (\sqrt{11})^2}\) \(f = 5\)
atleast i was close haha
finished ?? :p
i think so
cool!! well done
thnx for your help
mp
\(\color{blue}{\text{Originally Posted by}}\) @Tazmaniadevil ok she gave me a diffrent formula soo what do i do im so confused now \(\color{blue}{\text{End of Quote}}\) the formula is same difference is f and c variables f= foci \[\huge\rm f = \sqrt{ a^2 -b^2}\] is same as \[\huge\rm c^2= a^2-b^2 ~~~~~c=\sqrt{a^2-b^2}\]
a^2 and b^2 are already given so would be easy you write it as c= sqrt{36-11} :=)
OK

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