Question

The polar graph of the equation \[r = \frac{1}{2 - \cos\theta}\] has what shape? (Note: THis is supposedly Algebra II, so it's probably very basic polar. Therefore, I'm not very familiar with with polar coordinates and graphs.)

Answer 1

Plot out what happens when theta = 0, pi/2, pi, 3pi/2.

What does it look like?

Answer 2

Once you have an intuitive feel for what this is, you can work out the algebra, by substituting r = sqrt(x^2+y^2), cos theta = x/r

Answer 3

Well, isn't it a repeating parabola?

Answer 4

No.

Answer 5

for theta = 0, r = 1
theta = pi/2, r = 1/2
theta = pi, r = 1/3
theta = 3pi/2,r = 1/2

Sketch out those four points on the plane.

Answer 6

I'm new to polar, so I had to get something to graph it for me. Isn't there a special graphing system for polar coordinates?

Answer 7

Can you link me to something that's very basic (something that can teach me very basic polar) because I doubt I'll need to know much until Pre-Calculus

Answer 8

Seriously begin with pen and paper with the four points above. What are the location of those four points in Cartesian coordinates?

theta = 0, r = 1
Ths point lies along the direction of the positive x-axis, 1 from the origin. Hence its Cartesian coordinates are (1,0).

Now you figure out the Cartesian coordinates for these three points:
theta = pi/2, r = 1/2
theta = pi, r = 1/3
theta = 3pi/2,r = 1/2

...and then plot all four points.

Answer 9

THis is the sort of graph I keep on getting.

Answer 10

No. What are the coordinates of
theta = pi/2, r = 1/2?

This is in the positive y direction. Hence another point is (0, 1/2)

So two points are
(1,0)
and
(0,1/2)

What are the other two?

Answer 11

Is another point (0, 1/3)? I'm seeing if I'm on the right track.

Answer 12

theta = pi is in the *negative* y direction.

So (0, 1/3) isn't quite right.

Answer 13

So (0, -1/3)

Answer 14

**correction, the negative x direction.

Answer 15

theta = pi is the negative x direction, hence the third point is at
(-1/3, 0)

Now where's the fourth point?

Answer 16

theta r (x,y)
0 1 (1, 0)
pi/2 1/2 (0, 1/2)
pi 1/3 (-1/3, 0)
3pi/2 1/2 ....

Answer 17

(0, -1/2 ) Is this right?

Answer 18

Yes. Now, plot those four points and tell me what kind of geometric figure would go through all four of them.

Answer 19

I got an ellipse? These are the points I graphed.

Answer 20

Yes, an ellipse.

Answer 21

polar coordinates to rectangular coordinates
\[x=r \cos \theta \\ r=\sqrt{x^2+y^2} \\ you \ \ have \ \ r=\frac{1}{2-\cos \theta} \\ 2r-r \cos \theta=1 \ \ then \ \ 2\sqrt{x^2+y^2}-x=1 \\ 2\sqrt{x^2+y^2}=x+1 \\4(x^2+y^2)=(x+1)^2\\ 4y^2=-3x^2+2x+1\\4y^2=-3(x^2-\frac{2}{3}x)+1\\4y^2=-3(x^2-\frac{2}{3}x+\frac{1}{9}-\frac{1}{9})+1\\4y^2=-3(x^2-\frac{2}{3}x+\frac{1}{9})+\frac{1}{3}+1\\4y^2+3(x-\frac{1}{3})^2=\frac{4}{3} \\ 3y^2+\frac{9}{4}(x-\frac{1}{3})^2=1\\ its \ \ an \ \ Ellipse \]

Answer 22

Alright. Thanks! I didn't know how to convert polar coordinates to cartesian coordinates, so I just used the Unit Circle for this. Am I supposed to do that or is there a different process?

Answer 23

What is the definition of polar coordinates?

Answer 24

Isn't it a coordinate that represents the distance from a fixed point?

Answer 25

Not quite.

You should look this up in your text book and make sure you really understand it.

Answer 26

Well, I don't have a text book because this is for an Algebra II math team test. I'm guessing to challenge people they added this question in because I don't think I'm supposed to know much about polar coordinates and are giving props to those that do for some reason. Anyway, thank you!

Answer 27

Polar coordinates are defined as the distance from the origin, that is the variable r.

And angle to the positive x axis in the anticlockwise direction, this is the variable theta.

Answer 28

Draw yourself a diagram and convince yourself that
x = r cos(theta)
y = r sin(theta)

Answer 29

I'm finding this very comparable to the unit circle except r isn't always 1.

Answer 30

From this it follows that
\[ x^2 + y^2 = r^2 (\cos^2 \theta + \sin^2 \theta) = r^2 \]
i.e.,
\[ r =\sqrt{x^2 + y^2}\]

Answer 31

Ok. Thank you!

Answer 32

Also
\[ \frac{y}{x} = \frac{r \sin \theta}{r \cos \theta} = \tan \theta \]
Hence
\[ \theta = \arctan(y/x) \]