Question

help me

Answer 1

@LynFran

Answer 2

is it horizontal or vertical ?

Answer 3

??

Answer 4

is it horizontal ellipse or vertical ellipse?

Answer 5

idk lol

Answer 6

c^2=a^2-b^2

Answer 7

if bigger number under the x coordinate then horizontal
and if it's under the y coordiante then vertical

Answer 8

so its horizontal

Answer 9

yes 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

Answer 10

(h,k) is the center

Answer 11

ok

Answer 12

btw standard form of ellipse is \[\huge\rm \frac{ (x-h)^2 }{ a^2 } + \frac{ (y-k)^2 }{ b^2 }=1\]
a^2= ??
b^2= ??
in ur question

Answer 13

yeah how do i found this out

Answer 14

they already in the question
a^2 is under the x
b^2=under the y

Answer 15

ok sooooo

Answer 16

substitute a^2 and b^2 value into the equation lynn gave you
c^2=a^2-b^2 then solve for c

Answer 17

so hard im just now learing this lol

Answer 18

ohh

Answer 19

thats fine \[\huge\rm \frac{ (x-h)^2 }{ \color{ReD}{a^2 }} + \frac{ (y-k)^2 }{\color{ReD}{ b^2 }}=1\]
\[\huge\rm \frac{ x^2 }{\color{Red}{ 36} }+\frac{ y^2 }{\color{ReD}{ 11} }=1\]
bg number would be `a^2` always!(in ellipse equation)

Answer 20

so a^2 = ?

Answer 21

1

Answer 22

no
a^2 is under the x^2 coordinate

Answer 23

http://prntscr.com/89k6q9

Answer 24

so36

Answer 25

yes right now what is b^2=?

Answer 26

11

Answer 27

perfect!
\[\huge\rm c^2=a^2-b^2\]
substitute a^2 and b^2 for their values solve for c

Answer 28

c^2=-1175

Answer 29

am i right

Answer 30

ayooo
how :o

Answer 31

a^2=36
b^2=11
it's already squared you don't need to square them again

Answer 32

sooo

Answer 33

so.. substitute a^2 for 36
and b^2 for 11

Answer 34

25

Answer 35

so c^2=25

Answer 36

yes right now take square root both sides to cancel out the square

Answer 37

so c=5

Answer 38

yes right now what's the center point ?

Answer 39

idk

Answer 40

\[\huge\rm \frac{ (x-\color{reD}{h})^2 }{ a^2 } + \frac{ (y-\color{ReD}{k})^2 }{ b^2 }=1\] are there any h and k value at the numerator ?

Answer 41

??

Answer 42

look at the original equation is there any number with x ?

Answer 43

yeah its sqared

Answer 44

not square
(x-3)^2 then h would be 3
for this equation center point is (0,0)

Answer 45

so formula for foci is \[\huge\rm (h \pm c,k)\] where (h,k) is the center (0,0)
c=5
substitute variables for values