How do you graph #r=2sec(theta+45^circ)#?

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Answer 1 · 2018-07-03T15:52:23

See graph and explanation.

Use #( x, y ) = r ( cos theta, sin theta )#.

Here,

2 = #r cos (theta + pi/4 ) #

#= r ( cos theta cos (pi/4) - sin theta sin (pi/4))#

# =1/sqrt2( x - y )#, giving

#x - y = 2 sqrt 2#.

Note that the general polar equation of a straight line is

#r = p sec (theta - alpha)#, with polar #( p, alpha )# as the foot of

the altitude, from the pole, upon the straight line.

Here, the foot of the perpendicular is #( 2, -pi/4)#. See this plot on

the graph. In Cartesians, this is #( sqrt 2, - sqrt 2 )#

graph{(x - y - 2 sqrt 2)((x-1.414)^2+ (y+1.414)^2-0.01) = 0}