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

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3is it horizontal or vertical ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3is it horizontal ellipse or vertical ellipse?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3if bigger number under the x coordinate then horizontal and if it's under the y coordiante then vertical

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3yes right so thats mean you should add c into x coordinate of the center \[\huge\rm (h \pm c,k)\] now use that equation lynn gave u to find c value

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3btw standard form of ellipse is \[\huge\rm \frac{ (xh)^2 }{ a^2 } + \frac{ (yk)^2 }{ b^2 }=1\] a^2= ?? b^2= ?? in ur question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah how do i found this out

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3they already in the question a^2 is under the x b^2=under the y

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3substitute a^2 and b^2 value into the equation lynn gave you c^2=a^2b^2 then solve for c

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so hard im just now learing this lol

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3thats fine \[\huge\rm \frac{ (xh)^2 }{ \color{ReD}{a^2 }} + \frac{ (yk)^2 }{\color{ReD}{ b^2 }}=1\] \[\huge\rm \frac{ x^2 }{\color{Red}{ 36} }+\frac{ y^2 }{\color{ReD}{ 11} }=1\] bg number would be `a^2` always!(in ellipse equation)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3no a^2 is under the x^2 coordinate

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3yes right now what is b^2=?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3perfect! \[\huge\rm c^2=a^2b^2\] substitute a^2 and b^2 for their values solve for c

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3a^2=36 b^2=11 it's already squared you don't need to square them again

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3so.. substitute a^2 for 36 and b^2 for 11

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3yes right now take square root both sides to cancel out the square

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3yes right now what's the center point ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3\[\huge\rm \frac{ (x\color{reD}{h})^2 }{ a^2 } + \frac{ (y\color{ReD}{k})^2 }{ b^2 }=1\] are there any h and k value at the numerator ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3look at the original equation is there any number with x ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3not square (x3)^2 then h would be 3 for this equation center point is (0,0)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3so formula for foci is \[\huge\rm (h \pm c,k)\] where (h,k) is the center (0,0) c=5 substitute variables for values
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