Please help How do you graph the polar equation r=8-sec#theta# ?

1 Answer
A. S. Adikesavan
Aug 10, 2018

See the explanation and graphs.

Explanation:

As sec values #notin ( - 1, 1 )#,

#r = 8 - sec theta notin ( -1+ 8, 1 + 8 ) = ( 7, 9 )#

r is periodic,with period #2pi#.

Short Table, for $theta in #theta in [ 0, pi ]#, sans asymptotic

#pi/2 and 3/2pi#:

#( r, theta )#: ( 7, 0 ) ( 6.845, pi/6 ) ( 6, pi/3 ) ( oo, pi/2 )#

#( 9.155, 5/6pi ) ( 10, 2pi/3 ) ( 9, pi )#.

#r = 0, at theta =1.4455# rad.

The graph is symmetrical about #theta = 0#.

graph is outside the annular ( circular ) region # 7 < r < 9 ).

Converting to the Cartesian form, using

#( x, y ) = r ( cos theta, sin theta ) and 0 <= sqrt ( x^2 + y^2 ) = r#,

# (x^2 + y^2)^0.5(1+1/x) = 8 #,

The Socratic graph is immediate, with asymptote x = 0.
graph{((x^2 + y^2)^0.5(1+1/x) - 8)(x+0.01y)=0[-40 40 -20 20]}
See the bounding circles #r = 7 and r= 9#.
graph{((x^2 + y^2)^0.5(1+1/x) - 8)(x^2+y^2-49)(x^2+y^2-81)=0[-20 20 -10 10]}

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