How do you graph $r=4/(1-costheta)$ ?

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Answer 1 · 2016-05-26T03:53:18

This is the polar equation of the parabola with vertex at $V (2, 0))$ and focus S at the pole(r=0). The directrix is along $r cos theta = 4$

The equation is obtained from the definition of the parabola, 'the

distance from the focus = the distance from the directrix, referred

to the focus as pole r = 0 and the focus-to-vertex axis of the

parabola as the initial line,

$theta=0$ ( negative x-axis ).

The standard form is

$(semi latus rectum)/r = 1 + cos theta.$ .

Here, the initial line is reversed to make it

$(semi latus rectum)/r = 1 - cos theta.$ ..

After converting to Cartesian frame as $sqrt(x^2 + y^2) = x + 4$ , the

parabola is drawn, using Socratic graphic facility. The focus is at O.

See the directrix x + 4 = 0. .

graph{((x^2+y^2)^0.5-x-4)(x+4)=0}#

How do you graph r=4/(1-costheta) r=4/(1-costheta)?