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

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Answer 1 · 2018-08-11T14:32:34

$0 \leq r = 3 sec θ + 5 csc θ$ $0 <= r = 3 sec theta + 5 csc theta$

$= \frac{3 sin θ + 5 cos θ}{sin θ cos θ}$ $= ( 3 sin theta + 5 cos theta )/( sin theta cos theta )$

$= \sqrt{34} \frac{cos (θ - α)}{sin θ cos θ}$ $= sqrt 34 ( cos ( theta - alpha ))/( sin theta cos theta )$ ,

$α = arcsin (\frac{3}{\sqrt{34}})$ $alpha = arcsin( 3/sqrt34 )$

$r = 0$ $r =0$ , at $θ = 2 k π \pm α, k = 0, \pm 1, \pm 2, \pm 3, . .$ $theta = 2kpi +- alpha, k = 0, +-1, +-2, +-3, ..$

Zeros of $sin θ and cos θ$ $sin theta and cos theta$ ,

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

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] }

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