SOLUTION: Determine the eccentricity, the type of conic, and the directrix. r = (9)/(5 +6 cos θ)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Determine the eccentricity, the type of conic, and the directrix. r = (9)/(5 +6 cos θ)      Log On


   

Question 1140692: Determine the eccentricity, the type of conic, and the directrix.
r = (9)/(5 +6 cos θ)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

given:
r+=+9%2F%285+%2B6cos%28theta%29%29+
find:
the type of conic
the eccentricity
the directrix:

Standard form for conics in polar equations is
r+=%28ep%29%2F%281+%2B-+e%2Acos%28theta%29%29
where e is the eccentricity, the directrix is x = ±p if cosine

in your case:
r+=+9%2F%285+%2B6cos%28theta%29%29 multiply the numerator and denominator by 1%2F5
r+=+9%281%2F5%29%2F%285%281%2F5%29+%2B%281%2F5%296cos%28theta%29%29
r+=+%289%2F5%29%2F%281+%2B%286%2F5%29cos%28theta%29%29+

the eccentricity:
e=6%2F5
the directrix:
since ep=9%2F5=>p=%289%2F5%29%2Fe...substitute e
=>p=%289%2F5%29%2F%286%2F5%29
=>p=%289%2F6%29
=>p=%283%2F2%29
so, the directrix is x+=+p+=> x=3%2F2

the type of conic:
since e%3E1, you have +hyperbola