Question

@jim_thompson5910
mathhh

Answer 1

49x^2 + y^2 = 49

49x^2/49 + y^2/49 = 49/49

x^2 + (y^2)/49 = 1

(x-0)^2 + ((y-0)^2)/49 = 1

((x-0)^2)/(1^2) + ((y-0)^2)/(7^2) = 1

Answer 2

So this is an ellipse

center: (0,0)

Length of minor axis: 2*1 = 2 units

Length of major axis: 2*7 = 14 units

Answer 3

c^2 = a^2 - b^2

c^2 = 7^2 - 1^2

c^2 = 49 - 1

c^2 = 48

c = sqrt(48)

c = 4*sqrt(3)

This is the focal length

Answer 4

So one focus is at ( 0, 4*sqrt(3) ) and another focus is at (0, -4*sqrt(3))

Answer 5

thats long 0.0 lol okay lemme write this down and catch up on it :P

Answer 6

okay i got it now..

Answer 7

ok great

Answer 8

ok first we need the center

Answer 9

how do we find this?

Answer 10

uhmm, is it (0,0)?

Answer 11

the two vertices are (2, 5) and (2, -5)

Answer 12

the midpoint of this segment is ( (2+2)/2, (5+(-5))/2 ) = (2, 0)

Answer 13

So the center is actually (2, 0)

Answer 14

ohhh

Answer 15

so (h,k) = (2, 0) ---> h = 2, k = 0

Answer 16

okay got it

Answer 17

the focal distance is 8 units since (2, 8) is 8 units from the center (2, 0)

Answer 18

so c = 8

Answer 19

alright. got that.

Answer 20

a = 5 since 'a' is the distance from the center to either vertex

Answer 21

a^2 + b^2 = c^2

5^2 + b^2 = 8^2

25 + b^2 = 64

b^2 = 64 - 25

b^2 = 39

b = sqrt(39)

Answer 22

So we found this

h = 2
k = 0
a = 5
b = sqrt(39)

Answer 23

alright im following so far.

Answer 24

plug all this info into


( (y-k)^2 )/(a^2) - ( (x-h)^2 )/(b^2) = 1

Answer 25

do i need to find x and y?

Answer 26

no

Answer 27

those are variables that have no fixed value

Answer 28

good, and you can rewrite (y-0)^2 as just y^2, but that's optional

Answer 29

okk cool (: i have 2 more

Answer 30

ok

Answer 31

can you just quickly go through these cause i gotta go leave for work in 9 minutes.

Answer 32

oh well we can do this later then

Answer 33

I'm still tracking down the formulas we'll need

Answer 34

sound good?

Answer 35

nooo i gotta get this done :(

Answer 36

ok

Answer 37

Here's how I think it should be done, but don't have enough time to confirm

Answer 38

no need to confirm. it'll be fine

Answer 39

x + 5y^2 = 0

x = -5y^2

(x - 0) = -5(y-0)^2

(-1/5)(x - 0) = (y-0)^2

-4(1/20)(x - 0) = (y-0)^2

p = 1/20 ---> focal length is 1/20

Answer 40

The vertex is (0,0) since h = 0 and k = 0

Answer 41

The directrix is y = 1/20 since you add 1/20 to the x coordinate of the center

Answer 42

Oh sorry, I meant to say x = 1/20 instead of y = 1/20

Answer 43

The focus is (-1/20, 0) since you go 1/20 units to the left of the center

Answer 44

okay makes sense.. last one

Answer 45

how much time do we have?

Answer 46

about 5 minutes.

Answer 47

ok then I'll make it quick

Answer 48

kk

Answer 49

multiply every term by the LCD 2x to get

3*2x = 6x

(3/x)*2x = 6x/x = 6

(1/2)*2x = 2x/2 = x

(1/x)*2x = 2x/x = 2

-------------------------------------------------------

So the expression simplifies to



6x - 6
------
x - 2

Answer 50

that makes sense!

Answer 51

thank you!!!

Answer 52

ok awesome, yw

Answer 53

Off to work!! ttyl (:

Answer 54

cya later