Answer:
a) 10/3
b) hyperbola
c) x = ± 6/5
Step-by-step explanation:
a) A conic section with a focus at the origin, a directrix of x = ±p where p is a positive real number and positive eccentricity (e) has a polar equation:
[tex]r=\frac{ep}{1\pm e*cos\theta}[/tex]
Given the conic equation: [tex]r=\frac{12}{3-10cos\theta}[/tex]
We have to make it to be in the form [tex]r=\frac{ep}{1\pm e*cos\theta}[/tex]:
[tex]r=\frac{12}{3-10cos\theta}\\\\multiply\ both\ sides\ by\ \frac{1}{3} \\\\r=\frac{12*\frac{1}{3}}{(3-10cos\theta)*\frac{1}{3}}\\\\r=\frac{12*\frac{1}{3}}{3*\frac{1}{3}-10cos\theta*\frac{1}{3}}\\\\r=\frac{4}{1-\frac{10}{3}cos\theta } \\\\r=\frac{\frac{10}{3}(\frac{6}{5} ) }{1-\frac{10}{3}cos\theta }[/tex]
Comparing with [tex]r=\frac{ep}{1\pm e*cos\theta}[/tex]
e = 10/3 = 3.3333, p = 6/5
b) since the eccentricity = 3.33 > 1, it is a hyperbola
c) The equation of the directrix is x = ±p = ± 6/5
