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

Eccentricity = 5/6

Type of conic section; Ellipse

Directrix; x = -11/5

Step-by-step explanation:

The first step would be to write the polar equation of the conic section in standard form by multiplying the numerator and denominator by 1/6;

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

The eccentricity is given by the coefficient of cos theta in which case this would be the value 5/6. Therefore, the eccentricity of this conic section is 5/6.

The eccentricity is clearly between 0 and 1, implying that the conic section is an Ellipse.

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 = 11/6

5/6*d = 11/6

d = 11/5

Since the denominator has a minus sign then the ellipse opens towards the right and thus the equation of its directrix is;

x = -11/5

step-by-step explanation:

20% of 120 is 24. if you subtract 24 from 120 you get 96

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.