Question

write the standard form equation of an ellipse whose vertices on the major axis are (-2,2) and (8,2) and a focus at (7,2)
PLASE I NEED HELP

Answer 1

we need to find the center point of the ellipse. can you do that?

Answer 2

i only know that y=2

Answer 3

Standard form: (x-h)^2/a^2 + (y-k)^2/b^2=1

Answer 4

what is the middle between -2 and 8 ?

Answer 5

4

Answer 6

lets pretend it 5 :)

Answer 7

ok so it is 5 lol

Answer 8

wait, add and divide ...

Answer 9

8-2 = 6 ... /2 = 3

lets go with 3

Answer 10

ok :)

Answer 11

so our center is (3,2) we can use this in the general setup that katra provided right? how would that look? and by the way, are there any options to pick from?

Answer 12

no options and ok

Answer 13

(x+3)^2 + (y-2)^2

Answer 14

-3 nd -2 are better

(x-3)^2 + (y-2)^2
-------- + ------- = 1
a^2 b^2

Answer 15

why -3?

Answer 16

becasue when x=3, we want that part to zero out

Answer 17

ok

Answer 18

c is the distance from center to focus
a is the distance from center to vertex

ive got the picture bad, ill have to edit it

Answer 19

ok

Answer 20

now can you use the pythagorean thrm setup to solve for a and b?

Answer 21

im confused lol

Answer 22

confused on what?

Answer 23

how do i find the pythag if i dont know the letters

Answer 24

thats why i drew the picture ... and deifned that parts

Answer 25

so b is 3

Answer 26

a is equal to the distance from center to vertex, what is a?
c is equal to the distance from center to focus ...

b is 3 yes

Answer 27

c is 7

Answer 28

c is 4, but immaterial since we only really need to determine a and b

Answer 29

a = the distance from 3 to 8. so 5

Answer 30

oh c is 5

Answer 31

the confusion might be that the name of the parts of the ellipse are not quite in the same order as the abc parts in the pythag thrm

Answer 32

yes because the answer is (x-3)^2/25 + (y-2)^2/9=1

Answer 33

(x-3)^2 + (y-2)^2
-------- + ------- = 1
a^2 b^2

since a=5, and b=3 ... fill them in

Answer 34

25 and 9