Question

Hi everyone! :o) When graphing the parametric r=6/(1+2sin(theta)), i get points (6,0),(2,pi/2),(6,pi), and (-6,3pi/2)...if the answer is a hyperbola, how do you generate more points on the upper curve? I can't seem to figure out why the answer is supposed to be obvious when I can't seem to generate any more points on the upper curve which would then visually tell me it an obvious graph of a hyperbola! With only the points given above, I get a downward facing parabola with a single point above at (-6,3pi/2)...can anyone help clear this up for more? thanks!

Answer 1

How am I supposed to know that this graph is really a hyperbola without more points on the upper curve?

Answer 2

I think negative angles should give the other curve

Answer 3

going with your assumption then, (-6,-pi/2) gives the exact same point as (-6,3pi/2)

Answer 4

since the points keep going over the top of each other, i still can't seem to generate more points on the upper curve...any other ideas?

Answer 5

maybe if i use other angles like pi/4 or pi/3 or pi/6...maybe those would generate the points?

Answer 6

yes, and (-6 , 270º) is the same as (6,90º) (I think better in degrees)

if you use -60º (or 300º) you get
r= 6/(1+2*sin(300)) = 6/-0.7321= -8.196

the x coord will be r cos(300) = -4.1
y coord = r sin(300)= 7.1

Answer 7

ok so just to clarify...if I expand the angles that I am using to include other angle like pi/3 or -pi/3, the point will eventually show up and I will be able to tell that this grqaph is a hyperbola then right?

Answer 8

yes.
plot (r cos(A), r sin (A) ) for each A in the range 0 to 2pi

Answer 9

one other quick question phi if you can help me...

Answer 10

if you have a parametric equation like r=6/(5+5sin(theta)), you get points (6/5,0), (6/10,pi/2),(6/5,pi), and this wierd one (undefined, 3pi/2)...what on earth do you do when a point is undefined?

Answer 11

as you approach the angle 210º or 330º the denominator 1+2sin(A) approaches 0
that makes r approach infinity. the lines through 0,0 at those angles are the asymptotes

Answer 12

well what's wierd is that when I graph it, it is simply a downward facing parabol...and the graph doesn't show any asymptotes

Answer 13

maybe my calculator is being stupid...wolfram shows a series of upward facing parabolas with a bunch of asymptotes...hmmm

Answer 14

Here is a quick program

A= 211: 329;

r= 6./(1+2*sind(A))
x= r.*cosd(A);
y= r.*sind(A);
plot(x,y,'.')
shg

that plots the upper curve of the hyperbola

Answer 15

here is the lower limb

A= -29:208
r= 6./(1+2*sind(A))
x= r.*cosd(A);
y= r.*sind(A);
plot(x,y,'.')
shg

Answer 16

hey phi...thanks...thats great! one last question...are all these graphs being done in cartesian or polar graphs? silly question i know but it gets a little confusing

Answer 17

I used rectangular coordinates (x,y) points, but you would get the same answer using angles and lengths (polar coords)