Question

Questions Below

Answer 1

will they be x parts or y parts?

Answer 2

For which one? Second one asks for the conjugate and transverse going through the x and y axis

Answer 3

youve only got one question posted in this ....

Answer 4

Oh? It shows up as both for me. heres one

Answer 5

And the other

Answer 6

id say we cover the one with the verts to start with ...

Answer 7

Sure, Im all eyes

Answer 8

to find the verts, we need to zero out either the x parts or the y parts ... and in a hyperbola only one of those makes any sense

Answer 9

so, which one should we zero out to work the verts?

Answer 10

The x?

Answer 11

correct, so what value of x will zero out the x parts?

Answer 12

Err the five?

Answer 13

yes :) so, knowing this, we have to determine the verts as: (5,y1) and (5,y2)

so lets zero out the xs to give us:

(y-8)^2/4 = 1

solving for y we get?

Answer 14

Oh, 10 and 6?

Answer 15

correct

Answer 16

It's actually easier than i thought, thanks, would i be able to solve the second one like that? Or is it completely different?

Answer 17

the second one is asking for lengths; but its got a similar feel to it

Answer 18

the transverse axis is the one with the verts on it ... so we would have to determine which parts (x or y) make sense for it again

Answer 19

since we have 4 different options for transverse axis, once we know it we are done

Answer 20

Oh, okay, so wee need a step from the other one

Answer 21

correct, which parts (x or y) make sense to keep to find the verts? and thereby the transverse axis ...

Answer 22

we should keep the x this time?

Answer 23

lets see:

-x^2 = 1

or

y^2 = 1

one of these things is just not doable with "real numbers"

Answer 24

oh -x^2

Answer 25

yeah :) so our verts and transverse axis relate to the y parts again

Answer 26

the length of the transverse axis is the distance between the verts; we can either figure out the verts like last time, or there is a shortcut

Answer 27

Can we try both ways?

Answer 28

since we are concerned with the y parts; the undery part is a value; say b^2, such that 2b is the length of our transverse axis ... lets try it that way and then confirm it with the longer method

Answer 29

Okay thanks

Answer 30

\[\frac{y^2}{b^2}\to \frac{y^2}{49}\]
b^2 = 49, b = sqrt49 = 7

2(7) = 14

Answer 31

Oh, i see

Answer 32

now, to go the longer route: we know that x=0 gets rid of the x parts leacing us with:\[\frac{y^2}{49}=1\]solve for the y values that make this work

Answer 33

so we'd set it up like the last one now?

Answer 34

correct

Answer 35

(y-49)^2/ = 1 I just confused myself we're supposed to divide by the 49 right? not subtract it

Answer 36

lets not change the nature of the equation :)

y^2 / 49 = 1

Answer 37

yeah xD and that would give us 7, -7

Answer 38

good, and the distance between -7 and 7 is ???

Answer 39

14 I get it now

Answer 40

now .... since i showed yo the under part relates to axis length, how would you say we could go about finding the conjugate?

our y parts gave us the transverse soo .. where would we look for the conjugate?

Answer 41

our x

Answer 42

yep

Answer 43

so under the x part is a happy little 36, can you tell me how we could determine the conjugate axis from this?

Answer 44

divide it by 3?

Answer 45

not quite; recall i posted that all we have to do is sqrt and double ...

sqrt(49) = 7, and 7 doubled is 14 for the transverse

sqrt(36) = 6, and 6 doubled is 12 for our conjugate

Answer 46

Ooooh, okay. I see it now thank you :)

Answer 47

good luck