How do you convert #r=6sectheta# into cartesian form?

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Answer 1 · 2016-07-13T20:57:08

#r = 6sec(theta) -> x=6#

To convert from polar to cartesian, we can make use the following:

#x = rcos(theta)#
#y = rsin(theta)#
#r^2 = x^2+y^2#

Since #r = 6sec(theta)#, we now have #x = underbrace(6sec(theta))_(r)cos(theta)#

We can simplify this expression even further since

#sec(theta) = 1/(cos(theta))#, so our expression becomes

#x = 6/cancel(cos(theta)) * cancel(cos(theta)) = 6#

So our final answer is #x = 6#.

We can check this by graphing, for example.
enter image source here

In this graph, #x=6# and #r = 6sec(theta)# both overlap because they are essentially the same.