Answer
(a) e = 10/3
(b) Hyperbola
(c) x = -6/5
(d) Image below:
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Work Step by Step
(a)
$$r = \frac {12}{3 - 10cos(\theta)} \div \frac{3}{3} = \frac {4}{1 - \frac {10} 3 cos(\theta)}$$
Based on the original equation:
$$\frac{ed}{1 - esin(\theta)}$$
$$e = 10/3$$
(b)
$e > 1 $, therefore, this equation represents a hyperbola.
(c)
The equation has a $-cos(\theta)$ and the focus is at (0,0), thus, the equation of the directrix must be in the "$x = -c$" pattern.
In this case: $c = d$:
$$ed = 4 \longrightarrow d = \frac{4}{e} = \frac{4}{10/3} = 12/10 = 6/5$$
$$x = -6/5$$
(d)
Plot the points where $\theta = 0$, $\theta = \pi/2$, $\theta = \pi$ and $\theta = 3\pi/2$
$$r = \frac {12}{3 - 10cos(0)} \longrightarrow (-\frac {12}{7}, 0)$$
$$r = \frac {12}{3 - 10cos(\pi/2)} \longrightarrow (4, \pi/2)$$
$$r = \frac {12}{3 - 10cos(\pi)} \longrightarrow (\frac {12}{13}, \pi)$$
$$r = \frac {12}{3 - 10cos(3\pi/2)} \longrightarrow ( 4, 3\pi/2)$$
Draw a hyperbola that passes through these points.
