Answer:
Hyperbola, horizontal directricx at a distance 4/3 units above the pole
Step-by-step explanation:
* The polar equation for a conic i
- For a conic with a focus at the origin, if the directrix is
x = ± p, where p is a positive real number, and the eccentricity
is a positive real number e, the conic has a polar equation
# r = ep/(1 ± e cosФ)
- For a conic with a focus at the origin, if the directrix is
y = ± p, where p is a positive real number, and the eccentricity
is a positive real number e, the conic has a polar equation
# r = ep/(1 ± e sinФ)
- For a conic with eccentricity e,
# if 0 ≤ e < 1, the conic is an ellipse
# if e = 1, the conic is a parabola
# if e > 1, the conic is an hyperbola
* Lets solve the problem
∵ 4 = r/(2 + 3sinФ)
- From the rule above
∴ Directrix is y = ± p
- Divide up and down by 2 to make the 1st term in the
bracket down = 1
∴ r = 2/(1 + (3/2)sinФ)
- Compare it with the rule
∴ e = 3/2 > 1
∴ The conic is hyperbola
∵ ep = 2
∴ p = 2 ÷ e = 2 ÷ 3/2 = 2 × 2/3 = 4/3
∴ Directrix is y = ± 4/3
* Now we can describe the graph
- Hyperbola, horizontal directricx at a distance 4/3 units above the pole
