Find a polar equation for the conic with its focus at the pole and the given vertex or vertices: A. Parabola with vertex at (1,-pi/2)
B. Ellipse with vertices at (2,0),(8,pi)
C. Hyperbola , e=2, x=1

A) r = 2/(1 - sinθ)

B) r = 16/(5 + 3 cosθ)

C) r = 2/(1 + 2 cosθ)

Step-by-step explanation:

A) We are given the parabola vertex as:

(r, θ) = (1, -π/2)

The vertex coordinates indicates that it's below the pole and thus the equation of the conic section is represented by;

r = ed/(1 - e sinθ)

Since the equation is a parabola, then the eccentricity is 1.

Thus,plugging in the relevant values;

1 = (1 × d)/(1 - (1 × sin-π/2))

1 = d/(1 - (-1))

1 = d/(1 + 1)

1 + 1 = d

d = 2

Therefore the general equation would be gotten by putting 2 for d and 1 for E.

Thus;

r = (1 × 2)/(1 - (1 × sinθ))

r = 2/(1 - sinθ)

B) vertices of Ellipse are given as;

(2,0) and (8, π)

The equation will be represented by;

r = ep/(1 + e cosθ)

Where e which is the eccentricity is given by: e = c/a

c = (8 - 2)/2

c = 6/2

c = 3

And, a = (8 + 2)/2

a = 10/2

a = 5

Thus,e = 3/5

Thus;

r = (3/5)p/(1 + (3/5) cosθ))

Multiply both numerator and denominator by 5;

r = 3p/(5 + 3 cosθ))

Now, using the vertex (2,0) and plugging it, we have;

2 = 3p/(5 + 3 cos0))

2 = 3p/(5 + 3(1))

2 = 3p/8

p = 16/3

Thus,the equation is;

r = 3(16/3)/(5 + 3 cosθ))

r = 16/(5 + 3 cosθ)

C) Since the directrix is x = p, then the function in the denominator would be cosine. This is because x = 4 which is greater than 0.

It means that there will be an addition sign in the denominator. We use the standard form of;

r = ep/(1 + e cosθ)

So, e = 2 and p = 1

Then;

r = (2 × 1)/(1 + 2 cosθ)

r = 2/(1 + 2 cosθ)

three-fourths of an hour

36÷48=0.75

0.75=

b

step-by-step explanation: