Question

Convert r = 4sin theta sec ^{2} theta to a rectangular equation

Answer 1

I see no equation...

Answer 2

Is there suppose to be one?

Answer 3

\[r \sin(\theta) \sec^2(\theta) \text{ is a polar expression }\]

Answer 4

\[\text{ use the following } \\ r^2=x^2+y^2 \\ r \sin(\theta)=y \\ r \cos(\theta)=x \\ \tan(\theta)=\frac{y}{x}\]

Answer 5

\[r = 4\sin \theta \sec ^{2} \theta \] is the correct equation... i am so sorry

Answer 6

hmm...
maybe we can try to multiply r on both sides as a first step

Answer 7

\[r^2=4 \cdot r \sin(\theta) \cdot \sec^2(\theta)\]
so we know we can replace r^2 with x^2+y^2
and rsin(theta) with y

Answer 8

then try to divide sec^2(theta) on both sides

Answer 9

and see if you can figure it out from there
i think it shouldn't look to bad :)

Answer 10

alright, give me one second to write it out!

Answer 11

okay, but what do i do with that sec^2 theta on the bottom?

Answer 12

1/sec^2(theta)=cos^2(theta)

Answer 13

\[r^2 \cos^2(\theta)=4 \cdot r \sin(\theta)\]

Answer 14

if x=rcos(theta)
then x^2=?

Answer 15

x^2 = rcos(theta)^2 ?

Answer 16

\[x^2=(r \cos(\theta)))^2\]
\[x^2=r^2 \cos^2(\theta)\]

Answer 17

and that is exactly what you have on the left of your equation

Answer 18

okay so dont change the r^2 to x^2 + y^2 then?

Answer 19

nope we are going to use that r^2 with the cos^2(theta) to right the left hand side has x^2

Answer 20

to write*

Answer 21

alrighty

Answer 22

and you know to replace the rsin(theta) with?

Answer 23

y ?

Answer 24

yep yep

Answer 25

okay so the ending converted equation is something like \[x ^{2} - 4y\]

Answer 26

x^2=4y

Answer 27

and i think that is what you meant and yes

Answer 28

oh okay, thank you so much

Answer 29

I hope everything made sense :)

Answer 30

yes! your help is very much appreciated

Answer 31

thanks