See my notes and let me know if you need further clarification.
I hope you enjoy the great weekend we have in store for us.
Sir,
Thanks for the clarification on the last question , r= 2/(1-cosTheta). I did a royal job on this one.
Had I set up my table like you described in earlier assignment I may have seen my problem. I don't know what I did to my calculator to yield such coordinates. I plugged in 2/(1-cos(2pi/3)) etc. Also, I thought (r) must equal the original in order to determine symmetry. I failed here too.
For Example:...
r=2/(1-cos(-Theta)) equals ....r=2/(1+cos theta) or so I thought. ......
r=2/(1-cos(-Theta)) is not the same equation as r=2/(1+cos theta). However, since cos(-theta) = cos(theta),
r=2/(1-cos(-Theta)) equals r=2/(1 - cos (theta) ), so replacing theta by -theta doesn't change the function.
I fully understand where you say, Since cos(-theta) = cos(theta), but thought the above was the way you test.
I went ahead and set up a table for Theta, Cos Theta , and R=2/(1-cos Theta) And was able to graph the parabola on its side.
When I tried solving the equation I was basing my equation of a parabola as x^2 etc thus my reason for the following y^4= x^2+2x+1 or x= -1 y=0 r=1.
This isn't far off. y^4= x^2+2x+1 is just the square of y^2 = x + 1, which would have its vertex at (0, -1).
The graph does indeed cross the x axis at (-1,0) but I'm sure this was a coincidence. I yielded x^2 + y^2 = 4 + 4 x + x^2, then (y^2)^2= (4x +4)^2