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(Q10) Determine the eccentricity, the type of conic, and the directrix for r = 11/ 6 - 5 cos theta.

(Q10) Determine the eccentricity, the type of conic, and the directrix for r = 11/ 6 - 5 cos-example-1

2 Answers

2 votes

Answer:

C

Explanation:

edge

User Vlad Khomich
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2 votes

Answer:

Choice C

eccentricity = 5/6

conic section; ellipse

directrix; x = -11/5

Explanation:

The first step is to write the polar equation in standard form by dividing both the numerator and denominator by 6;


r=((11)/(6) )/(1-(5)/(6)cos(theta))

The eccentricity of the conic section is the coefficient of cos theta, thus;

e = 5/6

Since e lies between 0 and 1, the conic section is an ellipse.

To determine the equation of the directrix, we use the equation;

e*d = 11/6

5/6 *d = 11/6

d = 11/5

Since the conic section opens towards the right, the equation of the directrix becomes;

x = -11/5

User Etty
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