Identify the conic whose equation is given.
1 answer:
Answer:
Hyperbola
Step-by-step explanation:
The polar equation of a conic section with directrix ± d has the standard form:
r=ed/(1 ± ecosθ)
where e = the eccentricity.
The eccentricity determines the type of conic section:
e = 0 ⇒ circle
0 < e < 1 ⇒ ellipse
e = 1 ⇒ parabola
e > 1 ⇒ hyperbola
Step 1. <em>Convert the equation to standard form </em>
r = 4/(2 – 4 cosθ)
Divide numerator and denominator by 2
r = 2/(1 - 2cosθ)
Step 2. <em>Identify the conic </em>
e = 2, so the conic is a hyperbola.
The polar plot of the function (below) confirms that the conic is a hyperbola.
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Answer:
(g-f) (-1)= sqrt(15)
(f/g)(-1)= 0
(g+f)(2)=sqrt(3)-3
(g*f)(2)=-3*sqrt(3)
Step-by-step explanation:
We have to eval the expressions given in the point indicated.
Lets start by the first equation
(g-f)(-1)= g(-1) - f(-1)= =
Now, lest continue with the others
(f/g)(-1)= f(-1)/g(-1)= (1-1)/sqrt(15)=0
(g+f)(2)=g(2)+f(2)=sqrt(3)-3
(g*f)(2)=g(2)*f(2)=sqrt(3)*(-3)=-3sqrt(3)