How do you graph $r = 2 sec (θ + 45^{\circ})$ $r=2sec(theta+45^circ)$ ?

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Answer 1 · 2018-07-03T15:52:23

See graph and explanation.

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

Here,

2 = $r cos (θ + \frac{π}{4})$ $r cos (theta + pi/4 )$

$= r (cos θ cos (\frac{π}{4}) - sin θ sin (\frac{π}{4}))$ $= r ( cos theta cos (pi/4) - sin theta sin (pi/4))$

$= \frac{1}{\sqrt{2}} (x - y)$ $=1/sqrt2( x - y )$ , giving

$x - y = 2 \sqrt{2}$ $x - y = 2 sqrt 2$ .

Note that the general polar equation of a straight line is

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

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

Here, the foot of the perpendicular is $(2, - \frac{π}{4})$ $( 2, -pi/4)$ . See this plot on

the graph. In Cartesians, this is $(\sqrt{2}, - \sqrt{2})$ $( sqrt 2, - sqrt 2 )$

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

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