How do you graph #r=(23)/(7-5sin theta)#?

Question

2336 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-21T08:02:03

See Socratic graph and explanation.

Use #l/r = 1 + e cos (theta-alpha)#, with e < 1, represents an ellipse

with a focus at the pole (0, 0) and major axis along #theta = alpha#,

The semi major axis #a = l / (1-e^2)#.

This can be reorganized to the form

#(23/7)/r=1+5/7cos (theta+pi/2)# revealing that the graph is the

ellipse with a focus S(0, 0). e = 5/7, #alpha = -pi/2# , a =

161/24 = 6.71, nearly, and .

semi minor axis

#b = sqrt (la) = sqrt((23/7)(161/24)) = 23/sqrt (24)# = 4.7, nearly.

A short Table for tracing the ellipse.

#(r, theta)#:

#(0, 23/7) (46/9, pi/6) ( 23/2, pi/2) (46/9, 5/6pi) (23/7, pi)#

#(46/19, 7/6pi) (23/12, 3/2pi) (46/19, 11/6pi) (23/7, 2pi)#

See a Socratic graph. Note that the major axis ( length 13.42 ) is

along y-axis.

graph{(x^2+y^2)^0.5-23/7 -5/7 y=0[-10 10 -3 12]}