Determine the eccentricity, the type of conic, and the directrix for r = 9/1+6 cos theta. (picture provided below)

Eccentricity = 6

Type of conic section; Hyperbola

Directrix; x = 3/2

Step-by-step 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

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

i think this could be a possible answer (1 / 84) -1

good luck! : d