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Determine the eccentricity, the type of conic, and the directrix for r = 9/1+6 cos theta. (Picture Provided Below)

Determine the eccentricity, the type of conic, and the directrix for r = 9/1+6 cos-example-1

2 Answers

5 votes

Answer: Option D

Eccentricity = 6

Type of conic section; Hyperbola

Directrix; x = 3/2

answered
User Clockworkgeek
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3 votes

Answer:

Eccentricity = 6

Type of conic section; Hyperbola

Directrix; x = 3/2

Explanation:

The given polar equation of the conic section is in standard form;

The eccentricity is given by the coefficient of cos theta in which case this would be the value 6. Therefore, the eccentricity of the this conic section is 6.

The eccentricity is clearly greater than 1, implying that the conic section is a Hyperbola.

Since the conic section is in standard form, the numerator is the product of eccentricity and the value of the directrix, that is;

e*d = 9

6*d = 9

d = 3/2.

Since the denominator has a plus sign then the hyperbola opens towards the left and thus the equation of its directrix is;

x = 3/2

answered
User Kevin Mei
by
8.0k points
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