SOLUTION: I don't understand how my book comes up witht these answers: If anyone can help, I'd greatly appreciate it! Convert from polar equation to rectangular equation: Number 1 r=4

Algebra ->  Trigonometry-basics -> SOLUTION: I don't understand how my book comes up witht these answers: If anyone can help, I'd greatly appreciate it! Convert from polar equation to rectangular equation: Number 1 r=4      Log On


   



Question 339645: I don't understand how my book comes up witht these answers: If anyone can help, I'd greatly appreciate it!
Convert from polar equation to rectangular equation:
Number 1
r=4/1+2Sin(theta)
Answer is: X squared - 3Ysquared + 16y - 16 = 0
Also, this one...they ask to change it to a polar equation.
x squared - y squared = 12.
The answer given is r squared = 12sec (2theta) I don't see how they get these???

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
I don't understand how my book comes up witht these answers: If anyone can help, I'd greatly appreciate it!
Convert from polar equation to rectangular equation:
Number 1

r=4%2F%281%2B2Sin%28theta%29%29

Draw this picture which defines x,y,r, and q
and all their relationships.



r=4%2F%281%2B2Sin%28theta%29%29

Clear of fractions:

r%281%2B2Sin%28theta%29%29=4

Replace any trig functions:

Refer to the picture above and you can see that you can 
replace Sin%28theta%29 by y%2Fr

r%281%2B2%2Aexpr%28y%2Fr%29%29=4

r%2B2y=4

Now replace the r by using the Pythagorean theorem r=sqrt%28x%5E2%2By%5E2%29
from the picture above.

sqrt%28x%5E2%2By%5E2%29%2B2y=4

Isolate the radical term:

sqrt%28x%5E2%2By%5E2%29=4-2y

Square both sides:

%28sqrt%28x%5E2%2By%5E2%29%29%5E2=%284-2y%29%5E2

x%5E2%2By%5E2=%284-2y%29%5E2

x%5E2%2By%5E2=%284-2y%29%284-2y%29%7D%7D%0D%0A%0D%0A%7B%7B%7Bx%5E2%2By%5E2=16-8y-8y%2B4y%5E2

x%5E2%2By%5E2=16-16y%2B4y%5E2

x%5E2-3y%5E2+%2B+16y+-16+=+0

-------------------------------

Also, this one...they ask to change it to a polar equation.

x%5E2+-+y%5E2+=+12

Refer to the picture above, write the expressions for the sine and
the cosine:

x%2Fr=cos%28theta%29,  y%2Fr=sin%28theta%29

Solve for x and y:

x=r%2Acos%28theta%29,  y=r%2Asin%28theta%29

Substitute these:

%28r%2Acos%28theta%29%29%5E2+-+%28r%2Asin%28theta%29%29%5E2+=+12

r%5E2%2ACos%5E2theta+-+r%5E2%2ASin%5E2theta+=+12

Factor out r%5E2 on the left

r%5E2%28Cos%5E2theta-Sin%5E2theta%29=12

Now we can recognize that what's in the parentheses is the right side 
of the identity Cos%282theta%29=Cos%5E2theta-Sin%5E2theta which you
should have memorized. So substituting:

r%5E2%28Cos%282theta%29%29=+12%0D%0A%0D%0ASolve+for+%7B%7B%7Br%5E2

r%5E2=12%2F%28Cos%282theta%29%29

Replace the Cos%282theta%29 by 1%2FSec%282theta%29

r%5E2=12%2F%281%2FSec%282theta%29%29%29

Invert and multiply:

r%5E2=12%2ASec%282theta%29%29%29

Edwin