How do you graph #r =3 sec (theta) + 5 csc (theta)#?

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Answer 1 · 2018-08-11T14:32:34

#0 <= r = 3 sec theta + 5 csc theta#

#= ( 3 sin theta + 5 cos theta )/( sin theta cos theta )#

#= sqrt 34 ( cos ( theta - alpha ))/( sin theta cos theta )#,

#alpha = arcsin( 3/sqrt34 )#

#r =0#, at #theta = 2kpi +- alpha, k = 0, +-1, +-2, +-3, ..#

Zeros of #sin theta and cos theta#,

#theta = k/2pi, k = 0, +-1, +-2, +-3,...#

make non-negative r infinite. Alternately, all these are in the

directions of the x and y axes.

The Cartesian form of the equation is

#r = 3r/x + 5r/y rArr 3/x + 5/y = 1#, when #r ne 0#.

This equation #( x - 3 )( y - 5 ) = 15# represents the rectangular

hyperbola, with center at ( 3, 5 ) and x = 3 and y = 5 as

asymptotes.

See graph,
graph{((y-5)(x-3)-15)(x-3+0.0001y)(y-5+0.0001x)=0[-17 23 -5 15] }