Question A.1: Graphing a Conic in Polar Form Graph r = 8/4 + 4 sin θ .
Graphing a Conic in Polar Form
Graph r = \frac{8}{4 + 4 \sin θ} .
ALGEBRAIC SOLUTION
Divide both numerator and denominator by 4 to obtain
r = \frac{2}{1+ \sin θ} .
Based on the preceding table, this is the equation of a conic with ep = 2 and e = 1. Thus, p = 2. Because e = 1, the graph is a parabola. The focus is at the pole, and the directrix is horizontal, 2 units above the pole.
The vertex must have polar coordinates (1, 90°). Letting θ = 0° and θ = 180° gives the additional points (2, 0°) and (2, 180°). See Figure 2.
As expected, the graph is a parabola, and it opens downward because the directrix is above the pole.
GRAPHING CALCULATOR SOLUTION
Enter
r_{1} = \frac{8}{4+ 4 \sin θ} ,
where the calculator is in polar and degree modes with polar coordinate displays. Figure 3(a) shows the window settings, and Figure 3(b) shows the graph.
Notice that the point (1, 90°) is indicated at the bottom.
