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anonymous
 one year ago
help me plz
anonymous
 one year ago
help me plz

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2first 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` \[\rm \frac{ (xh)^2 }{\color{ReD}{ a^2} }+ \frac{ (yk)^2 }{ b^2 }=1\] if a is under the y variable then it would be `vertical equation` \[\rm \frac{ (xh)^2 }{ b^2 }+ \frac{ (yk)^2 }{\color{reD}{ a^2} }=1\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2so is it vertical or horizontal ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think its vertical

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2foci for horizontal \[\large\rm (h, k \pm c)\] and for vertical equation \[\large\rm (h \pm c , k)\] use the equation to find c \[\huge\rm c^2 =a^2b^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so which one do i use

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2well a is bigger number which is under the x variable so is it vertical or horizontal ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2read my first comment :=)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it is horizontal right

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2yes right \[\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1\] now use the equation posted above to find c

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2foci 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 \[\huge\rm c^2 =a^2b^2\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2a^2 and b^2 are in the given equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow ok so umm can you start me off plz

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2`horizontal equation` for ellipse \[\rm \frac{ (xh)^2 }{\color{ReD}{ a^2} }+ \frac{ (yk)^2 }{ b^2 }=1\] the number at the denominator are a^2 and b^2 and like i said bigger number equal to a^2

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2what two numbers are replaced with a^2 and b^2 in this equation \[\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1\] ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0brb i gtg cook dinner

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2right which one is a^2 and which one is b^2 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0{ (xh)^2 }{36} }+/{ (yk)^2 }{11}=1 is this the correct formula

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2this ellipse is centred on the origin yep?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444085734817:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2Paraphrasing Wiki: The distance from the origin O to either focus is f where: \[f = \sqrt{a^2b^2}.\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok she gave me a diffrent formula soo what do i do im so confused now

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2here, what is a and what is b? general equation for ellipse centred on Origin is: \(\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2then dw:1444086259207:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(xh)^2 }{36} }+/{ (yk)^2 }{11}= ok so i got this

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so that means you have finished it?!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no i have not thats the first equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(yk)^2 {11} this is my answer

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2OK 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 \[f = \sqrt{a^2b^2}\] dw:1444086808815:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2soz, we crossed posts. i understood the question to be finding the focii of the ellipse

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes idk what im doing im so cunfused here im sorrry

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2no 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: \(\large f = \sqrt{a^2b^2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0omg i have no clue on how to do this like wth

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\(a = 6, \; b = \sqrt{11}\) \(f = \sqrt{ 6^2  (\sqrt{11})^2}\) \(f = 5\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0atleast i was close haha

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\(\color{blue}{\text{Originally Posted by}}\) @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 \[\huge\rm c^2= a^2b^2 ~~~~~c=\sqrt{a^2b^2}\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2a^2 and b^2 are already given so would be easy you write it as c= sqrt{3611} :=)
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