Question

Find the standard form of the equation of the ellipse and give the location of its foci.

Answer 1

please explain

Answer 2

Do you know how to find major axis and minor axis?

Answer 3

no ma'am

Answer 4

major axis= 2a, minor axis=2b

Answer 5

look at the graph

Answer 6

looking o.o

Answer 7

do you know where the major axis is?

Answer 8

half of the value of the major axis is a (we need a and b for the equation)

Answer 9

semi-major axis is 7 and the semi-minor is 6.

Answer 10

yes :)

Answer 11

7 is a and 6 is b

Answer 12

the standard equation for an ellipse ( in this case it is vertical ellipse because the major axis is in the y axis) \[\frac{ x^{2} }{ b^{2} } + \frac{ y^{2} }{ a^2 }=1\]

Answer 13

So I'm supposed to plug in the major axis and the minor axis into a and b? and if the major axis is in the x-axis it's a horizontal ellipse?

Answer 14

\(\Large \frac{ x^{2} }{ a^{2} } + \frac{ y^{2} }{ b^2 }=1 \)

You can quickly verify where 'a' and 'b' should be by setting y = 0 and finding the x-intercepts or setting x = 0 and finding the y-intercepts.

Answer 15

yes

Answer 16

I see...
do the x and y need numbers plugged into them too?

Answer 17

no

Answer 18

Plug in values for 'a' and 'b'.

Answer 19

can you tell us what you get for the equation?

Answer 20

when you plug in

Answer 21

The major axis is along x-axis so 'a^2' should go under the x-term.

Answer 22

\[\frac{ x^2 }{ 49 } + \frac{ y^2 }{ 36 } = 1\]

Answer 23

oops sorry about that

Answer 24

o.o dang it

Answer 25

yes that is correct

Answer 26

Ooooh lol sorry

Answer 27

but your equation is correct ^_^

Answer 28

thank you guys...what do I do next? o.o

Answer 29

ok now let's find the Foci

Answer 30

Foci of a Horizontal ellipse: (h+-c,k)

Answer 31

The vertex is (h,k), for this ellipse, its vertex is at the origin (0,0)

Answer 32

Woah there...sorry I just got confused e.e

Answer 33

oh wait nvm I see it

Answer 34

the foci of a horizontal ellipse: (h+-c,k)

the +- means plus or minus)

Answer 35

to find c you use: \[ c^{2}=a^{2}-b^{2}\]

Answer 36

\[c^2 = 7^2 - 6^2\] then shouldn't c^2 be 1^2 ?

but in my calc I did 49 - 36 and got 13... would I make that \[\sqrt{13}\]

Answer 37

yes

Answer 38

Yes. One focus at \(\large (+\sqrt{13}, 0) \) and another at \(\large (-\sqrt{13}, 0) \)

Answer 39

I agree with aum

Answer 40

FYI:
In a parabola (h,k) is the vertex.
In ellipse and hyperbola, (h,k) is the center.

Answer 41

correct,lol sorry, I am in pre-calculus hehe I'm learning as well,but thank you aum :)

Answer 42

Thank you both ^-^ I wish I could give you both medals /.\ I appreciate your help <3