Question

polar to rectangular

solve it in the standard form hyperbola

r=8/(2-5 cos (theta))

Answer 1

\[4y ^{2}-21(x ^{2}-\frac{ 80 }{ 21 }x + \frac{ 40 }{ 21 }-\frac{ 40 }{21})=64\]

got stuck at this part

Answer 2

Hmm before you go to complete the square I wanna see if you missed a negative sign a sec:\[\Large 4y^2-21x^2-80x\quad=\quad64\]\[\Large 4y^2-(21x^2+80x)\quad=\quad64\]

Answer 3

Did you get that same thing or no? :s

Answer 4

\[\frac{ (y-0)^{2} }{ (\sqrt{26})^{2} }-\frac{ (x+\frac{ 2\sqrt{210} }{ 21 })^{2} }{ (\sqrt{\frac{ 104 }{ 21 }})^{2} }=1\]

final answer

Answer 5

forgot i factored the negative.....

Answer 6

:'c

Answer 7

restart....

Answer 8

\[4y ^{2}-(-21(x ^{2}+80x))=64\]

Answer 9

looks right?

Answer 10

Looks like there is an extra negative.. hmm

Answer 11

4y2−(21(x2+80x))=64

like this?

Answer 12

Squaring gave you:\[\Large 4x^2+4y^2\quad=\quad 25x^2+80x+64\]Moving the x's to the left side,\[\Large 4y^2-21x^2-80x\quad=\quad64\]\[\Large 4y^2-21\left(x^2+\frac{80}{21}x\right)\quad=\quad64\]

Answer 13

When you factor out that 21, don't forget to pull it out of the 80 as well! :)
Like you did the first time *

Answer 14

ok i got that...
now completing the square

\[4y^2-21(x^2+\frac{ 80 }{ 21 }x +\frac{ 40 }{ 21 }-\frac{ 40 }{ 21 })=64\]

Answer 15

Woops, you took half of the b value, good... but you forgot to square it.

Answer 16

\[\frac{ 40 }{ 21 } --> \frac{ 1600 }{ 441 }\]

Answer 17

I don't think we want to square it.
I mean, I think we want to leave the exponent on it.
It'll make things a tad easier for us.

Answer 18

yeah... i wrote the exponent

Answer 19

\[4y ^{2}-21(x+\frac{ 40 }{ 21 })^{2}=64+21(\frac{ 40 }{ 21 })^2\]

Answer 20

I think you missed a negative, there are a bunch to keep track of when we move it outside of the brackets.\[\Large 4y^2-21\left(x^2+\frac{80}{21}x+\frac{40^2}{21^2}\color{#CC0033}{-\frac{40^2}{21^2}}\right)\quad=\quad64\]Taking the red term out of the brackets gives us:\[\Large 4y^2-21\left(x^2+\frac{80}{21}x+\frac{40^2}{21^2}\right)\color{#CC0033}{+\frac{40^2}{21}}\quad=\quad64\]Moving to the other side and writing our completed square:\[\Large 4y^2-21\left(x+\frac{40}{21}\right)^2\quad=\quad64\color{#CC0033}{-\frac{40^2}{21}}\]

Answer 21

then we will have a negative on the right hand side which im again stuck....

Answer 22

So on the right side, we'll get a common denominator:\[\Large 4y^2-21\left(x+\frac{40}{21}\right)^2\quad=\quad\frac{1344-1600}{21}\]
Is that the part you're stuck on? Or earlier than that?

Answer 23

that is the part

Answer 24

Subtracting gives us:\[\Large 4y^2-21\left(x+\frac{40}{21}\right)^2\quad=\quad-\frac{256}{21}\]To deal with the negative, we multiply each side by -1,\[\Large 21\left(x+\frac{40}{21}\right)^2-4y^2\quad=\quad\frac{256}{21}\]

Answer 25

do I multiply everything by \[\frac{21 }{ 256 }\] in order to get 1 on the right side ?

Answer 26

Ya that sounds right.

Answer 27

\[\frac{ y^2 }{ (\frac{ 8\sqrt{21} }{ 21 })^{2} }

Answer 28

... hard to type the final answer

Answer 29

\[\frac{ y^2 }{ (\frac{ 8\sqrt{21} }{ 21 })^{2} }+ \frac{ x+\frac{ 40^{2} }{ 21 } }{ \frac{ 16 }{ 21 })^{2} }= 1\]

Answer 30

the ( 40/21) should be (40/21)^2

Answer 31

Sorry I couldn't get openstudy to work..

\[\Large \frac{y^2}{\left(\cfrac{ \color{red}{16}\sqrt{21}}{21}\right)^{2} }+ \frac{ \color{red}{\left(\color{black}{x+\frac{40}{21}}\right)^2} }{ \left(\cfrac{16}{21}\right)^{2} }= 1\]

Answer 32

Check out my notes in red, I think you made minor mistake in those spots.

Answer 33

Oh also, shouldn't the y term be negative? :o

Answer 34

\[\Large \frac{ \color{red}{\left(\color{black}{x+\frac{40}{21}}\right)^2} }{ \left(\cfrac{16}{21}\right)^{2} }\color{red}{-}\frac{y^2}{\left(\cfrac{ \color{red}{16}\sqrt{21}}{21}\right)^{2}}=1\]

Answer 35

These problems are painful...
Do you have an answer key or some way to check our work?

Answer 36

nope... its a review question for my coming test on friday

Answer 37

the answers will be available to us on friday and these will be counted as 5% of our grade for this test

Answer 38

Oh interesting :O

Answer 39

why was it \[\frac{ 16\sqrt{21} }{ 21 }\]

Answer 40

Oh i did something stupid there.. lemme see...

Answer 41

since it is \[\sqrt{\frac{ 64 }{ 21 }}\]

Answer 42

rationalized should be\[\frac{ 8\sqrt{21} }{ 21 }\]

Answer 43

\[\Large \frac{y^2}{\left(\dfrac{16^2}{4\cdot21}\right)}\]The y term is something like that, right?

Answer 44

\[\Large \frac{4y^2}{\left(\dfrac{16^2}{21}\right)}\quad=\quad \frac{y^2}{\left(\dfrac{16^2}{4\cdot21}\right)}\]

Answer 45

Oh ok ok ok :) I'm slow lolol
Ya I see what you did there :3

Answer 46

im lost lol...

Answer 47

\[\Large \frac{y^2}{\left(\dfrac{16^2}{4\cdot21}\right)}\quad=\quad \frac{y^2}{\left(\dfrac{16}{2\sqrt{21}}\right)^2}\]Ok ok now it makes sense :D

Answer 48

Are you? :3 ah crap!

Answer 49

\[\Large \frac{ \color{red}{\left(\color{black}{x+\frac{40}{21}}\right)^2} }{ \left(\cfrac{16}{21}\right)^{2} }\color{red}{-}\frac{y^2}{\left(\cfrac{ \color{red}{8}\sqrt{21}}{21}\right)^{2}}=1\]
So uhh something like this? +_+

Answer 50

thats what i got

Answer 51

gonna have a "great time" graphing this

Answer 52

lol

Answer 53

thnx so much by the way

Answer 54

np c: