Question

convert from polar to rectangular

r=4/(5-3 sin (theta))

I got (25/16 X^2) + y^2 - (24/16 y) = 1
then i am stuck

need to be in standard form of ellipse ( im guessing ellipse but could not get the form right) or hyperbola

Answer 1

Should work out to an ellipse? Mmm ok let's see if we can figure this out...

Answer 2

I can see that it is an ellipse but i cant get it to be like the standard from as i need to graph it

Answer 3

\[\Large r\quad=\quad \frac{4}{5-3\sin \theta}\]So doing some multiplication:\[\Large 5r-3r \sin \theta\quad=\quad 4\]We can covert the r and r sin theta into some stuff,\[\Large 5\sqrt{x^2+y^2}-3y\quad=\quad 4\]Make sense so far? :O Look ok?

Answer 4

yup exactly what i got

Answer 5

Hmmm let's see..\[\Large 5\sqrt{x^2+y^2}\quad=\quad 4+3y\]Squaring:\[\Large 5x^2+5y^2\quad=\quad 9y^2+24y+16\]Then we have to simplify this down? :O
Mmm doesn't look too bad :D

Answer 6

I got (25/16 X^2) + y^2 - (24/16 y) = 1
then i am stuck

Answer 7

5x2+5y2 shouldn't the 5 be squared as well since we are suppose to square everything right?

Answer 8

Ah yes, good catch! ^^

Answer 9

Mmm ok lemme fix that a sec before I write anything else lol.\[\Large 25x^2+25y^2\quad=\quad 9y^2+24y+16\]

Answer 10

alright i got that too...

Answer 11

Ya lemme erase that garbage a sec :x let's do that a smarter way

Answer 12

Moving the y's to the left gives us:
\[\Large 25x^2+\color{#CC0033}{16y^2-24y}\quad=\quad16\]
Now we need to complete the square on the y's before we do anything else.

Answer 13

ok im doing it too hope it come out to the standard form

Answer 14

yeah i am stuck
\[25x ^{2}+16(y+\frac{ 3 }{ 4 })^{2}=16+\frac{ 9 }{ 16 }\]

Answer 15

\[\Large 25x^2+16\left(\color{#CC0033}{y^2-\frac{3}{2}y}\right)\quad=\quad16\]
\[\Large 25x^2+16\left(\color{#CC0033}{y^2-\frac{3}{2}y+\frac{9}{16}}\right)\quad=\quad16\color{#CC0033}{+9}\]
\[\Large 25x^2+16\left(\color{#CC0033}{y-\frac{3}{4}}\right)^2\quad=\quad25\]

Answer 16

So when you added 9/16 to complete the square, you should also subtract 9/16 `within the brackets`.
So to pull that value out, you need to multiply it by 16, right? Because of the 16 on the brackets?

Answer 17

oh yeah....

Answer 18

So I think then we just need to divide by 25, yes? :)

Answer 19

We might have to do a little work with the y term so it looks more "standard".

Answer 20

How do we get the x term to be \[(x-h)^{2}\]

Answer 21

\[\Large x^2+\frac{16\left(\color{#CC0033}{y-\frac{3}{4}}\right)^2}{25}\quad=\quad1\]
\[\Large x^2+\frac{\left(y-\frac{3}{4}\right)^2}{\left(\frac{5}{4}\right)^2}\quad=\quad1\]So like that for the y's maybe?

Answer 22

The x is already in standard form:\[\Large \frac{(x-0)^2}{1^2}\quad=\quad x^2\]See? :o

Answer 23

Okay now i see

Answer 24

Hmm I hope we did that correctly :) Bit of a tricky problem to simplify.

Answer 25

i checked everything as we progress should be fine thank you so much

Answer 26

cool c: