Question

Find the graphing form, coordinates of the center & foci, the length of the transverse and conjugate axis, then graph the points and equation. 16x2 -81y2 + 324 y – 1620 = 0

Answer 1

ok this one is not an ellipse is it?
do you know what it is and how to tell?

Answer 2

its asking for the length of the transverse and conjugate axis?

Answer 3

lol maybe that is one way
i had in mind that you have \(-81y^2\) not \(+81y^2\)

Answer 4

we are going to do the same thing as before
complete the square twice to make it look like
\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]

Answer 5

you think you can do that or you want me to walk you through it?

Answer 6

walk me through it, i still need more practice

Answer 7

ok
there is only \(16x^2\) with no other \(x\) term so it will be easier
factor out the \(-81\) from
\[-81y^2+384y\] what do you get?

Answer 8

\[16x^2-81(y^2+?)=1620\] what goes in the \(?\) spot? i.e. what is \(384\div(-81)\)

Answer 9

-4.7

Answer 10

whew i made a big mistake, i meant what is \(324\div (-81)\)

Answer 11

i copied it down wrong sorry

Answer 12

lol oh man, okay im confused now

Answer 13

it was just a stupid typo on my part, don't let that throw you
you wrote 324 and i copied it as 384

Answer 14

so its -81y^2 - 324y?

Answer 15

you have to factor out the \(-81\)

Answer 16

\[16x^2-81(y^2-4y)=1620\]

Answer 17

since \(-81\times -4=324\)

Answer 18

did i lose you or is that okay now?

Answer 19

i need you to get me back on track, what all i have written down is -81y^2 + 38y ,

Answer 20

what do i do now?

Answer 21

lets start from square one

Answer 22

so delete that then?

Answer 23

yeah

Answer 24

okay done

Answer 25

lets start with
\[16x^2 -81y^2 + 324y – 1620 = 0 \]add \(1620\) get
\[16x^2-81y^2+324y=1620\] that part is easy

Answer 26

now we are going to factor out the \(-81\) so we can complete the square

Answer 27

\[16x^2-81(y^2-4y)=1620\] let me know when that is clear

Answer 28

got it

Answer 29

ok good now to complete the square

Answer 30

half of 4 is 2, so we write
\[16x^2-81(y-2)^2=1620-?\]and we have to figure out what goes in the \(?\)

Answer 31

\(y-2)^2=y^2-4y+4\) multiply that \(4\) by \(-81\) and you get \(324\) so you have to subtract \(324\) from the right
\[16x^2-81(y-2)^2=1620-324\]

Answer 32

\[16x^2-81(y-2)^2=1296\] then divide

Answer 33

i have a feeling i might have lost you somewhere
let me know if that is true, and where

Answer 34

no you didnt i was just looking over what we're doing

Answer 35

ok finally divide both sides by \(1296\) to put a 1 on the right

Answer 36

this step always confuses me

Answer 37

you do get that the right side will be 1 yes?
when you divide by \(296\)?

Answer 38

1296

Answer 39

yes i got that part

Answer 40

then these are usually cooked up to cancel nicely
\[\frac{16}{1296}=\frac{1}{81}\]

Answer 41

so the first term will be
\[\frac{x^2}{81}\]

Answer 42

and likewise \(\frac{81}{1296}=\frac{1}{16}\) so the next term will be
\[-\frac{(y-2)^2}{16}\]

Answer 43

then standard form is
\[\frac{x^2}{81}-\frac{(y-2)^2}{16}=1\]

Answer 44

now you can read off the center from your eyeballs again

Answer 45

\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]\[\frac{x^2}{81}-\frac{(y-2)^2}{16}=1\]

Answer 46

\((h,k)=(0,2)\)

Answer 47

thats the center?

Answer 48

yes

Answer 49

what else you need?

Answer 50

foci and length of transverse and conjugate axis .-.

Answer 51

ok again we need to know something
does it look like the one on the left or on the right (lousy picture)