Question

PLEASE HELP

8.08, part 2

11. Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).

A) y squared over 45 minus x squared over 36 = 1
B) y squared over 81 minus x squared over 36 = 1
C) y squared over 36 minus x squared over 81 = 1
D) y squared over 36 minus x squared over 45 = 1

12. Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ± 1 divided by 4. x.

A) y squared over 16 minus x squared over 64 = 1
B) y squared over 16 minus x squared over 256 = 1
C) y squared over 256 minus x squared over 16 = 1
D) y squared over 64 minus x squared over 4 = 1

13. Eliminate the parameter.
x = t - 3, y = t2 + 5

A) y = x2 + 6x + 14
B) y = x2 - 14
C) y = x2 - 6x - 14
D) y = x2 + 14

14. Find the rectangular coordinates of the point with the polar coordinates.
ordered pair 3 comma 2 pi divided by 3

A) ordered pair negative 3 divided by 2 comma 3 square root 3 divided by 2
B) ordered pair 3 square root 3 divided by 2 comma negative 3 divided by 2
C) ordered pair negative 3 divided by 2 comma 3 divided by 2
D) ordered pair 3 divided by 2 comma negative 3 divided by 2

15. Find all polar coordinates of point P where P = negative pi divided by 6 .

A) (1, negative pi divided by 6 + (2n + 1)π) or (-1, negative pi divided by 6 + 2nπ)
B) (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + 2nπ)
C) (1, negative pi divided by 6 + 2nπ) or (1, pi divided by 6 + (2n + 1)π)
D) (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n + 1)π)

16. Determine two pairs of polar coordinates for the point (4, 4) with 0° ≤ θ < 360°.

A) (4 square root 2 , 135°), (-4 square root 2 , 315°)
B) (4 square root 2 , 45°), (-4 square root 2 , 225°)
C) (4 square root 2 , 315°), (-4 square root 2 , 135°)
D) (4 square root 2 , 225°), (-4 square root 2 , 45°)

17. The graph of a limacon curve is given. Without using your graphing calculator, determine which equation is correct for the graph.
a circular graph with an inner loop on the left


[-5, 5] by [-5, 5] (5 points)

A) r = 3 + 2 cos θ
B) r = 2 + 3 cos θ
C) r = 2 + 2 cos θ
D) r = 4 + cos θ

18. Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.
r = -2 + 3 cos θ

A) No symmetry
B) y-axis only
C) x-axis only
D) Origin only

19. A railroad tunnel is shaped like a semiellipse, as shown below.
A semiellipse is shown on the coordinate plane with vertices on the x axis and one point of intersection with the positive y axis.

The height of the tunnel at the center is 54 ft, and the vertical clearance must be 18 ft at a point 8 ft from the center. Find an equation for the ellipse.


20. Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.
r = 2 cos 3θ

Answer 1

11. Ans: (D)

Since all the vertices and the foci lie along the y axis, therefore, we would need the following equation for vertical hyperbola:


`((y-k)^2)/(a^2) - ((x-h)^2)/(b^2) = 1`

Since (h,k) = (0,0)
Therefore, the above equation becomes,

`((y)^2)/(a^2) - ((x)^2)/(b^2) = 1`

Now the distance between the vertices is:
2a = 12
=> a = 6

And the distance between the foci is:
2c = 18
=> c = 9

Since,

`c^2 = a^2 + b^2`

=>
`b^2 = 45`

Hence, the equation becomes,

`((y)^2)/(36) - ((x)^2)/(45) = 1` (Option D:y squared over 36 minus x squared over 45 = 1)

12. Ans: (B)
The hyperbola's standard form is(as it is a vertical):

`(y^2)/(16) - (x^2)/(b^2) = 1` -- (X)

=>
`y^2 = ( (16)/(b^2))*(b^2 + x^2)`

=> y = ±
`( (4)/(b) ).x` --- (A)
Since asymptotes at y = ±
`( (1)/(4) ).x`. --- (B)
Compare (A) and (B), you would get,

`(4)/(b) = (1)/(4)`

=> b=16

The equation (X) would become:

`(y^2)/(16) - (x^2)/(256) = 1` (Option-B)


13. Ans: (A)
`y = x^(2) + 6x + 14`
Equations given:
x = t - 3 --- (equation-1)
y =
`t^(2)` + 5 --- (equation-2)

From equation-1,
t = x + 3

Put the value of t in (equation-2),

`y = (x+3)^(2) + 5`

`y = x^2 + 9 + 6x + 5`

`y = x^2 + 6x + 14`

Hence, the correct option is (A)

14. Ans: (A)

The polar coordinates given:
`(3, (2 \pi )/(3) )` = (r, θ)
Since,
x = r*cosθ,
y = r*sinθ

Plug-in the values of r, and θ in the above equations:
x = (3) * cos(120°); since
`(2 \pi )/(3)` = 120°
=> x =
`- (3)/(2)`

y = (3) * sin(120°);
=> y =
`(3 √(3) )/(2)`

Ans: (x,y) =
`(- (3)/(2) ,(3 √(3) )/(2))` (Option A)

15. Ans: (D)
The general forms of finding all the polar coordinates are:
1) When r >= 0(meaning positive): (r, θ + 2n
`\pi`) where, n = integer
2) When r < 0(meaning negative): (-r, θ + (2n+1)
`\pi`) where, n = integer

Since r is not mentioned in the question, but in options every r slot has the value r=1, therefore, I would take r = +1, -1(plus minus 1)

θ(given) =
`(- \pi )/(6)`

When r = +1(r>0):
(1,
`(- \pi )/(6)` + 2n
`\pi`)

When r = -1(r<0):
(-1,
`(- \pi )/(6)` + (2n+1)
`\pi`)

Therefore, the correct option is (D): (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n + 1)π)

16. Ans: (B)
In polar coordinates,

`r = \sqrt{x^(2) + y^(2)}`

Since x = 4, y=4; therefore,

`r = √(16 + 16) = 4 √(2)`

To find the angle,
tanθ = y/x = 4/4 = 1

=> θ = 45° (when
`r =4 √(2)`)
If r = -
`r =4 √(2)`, then,

θ = 45° + 180° = 225°
Therefore, the correct option is (B) (4 square root 2 , 45°), (-4 square root 2 , 225°)

17. Ans: (B)

(Question-17 missing Image is attached below) The general form of the limacon curve is:
r = b + a cosθ

If b < a, the curve would have inner loop. As you can see in the image attached(labeled Question-17), the limacon curve graph has the inner loop. Therefore, the correct option is (B) r = 2 + 3 cosθ, since b = 2, and a = 3; and the condition b < a (2 < 3) is met.


18. Ans: (C)
Let's find out!
1. If we replace θ with -θ, we would get:
r = -2 + 3*cos(-θ )
Since, cos(-θ) = +cosθ, therefore,
r = -2 + 3*cos(θ)

Same as the original, therefore, graph is symmetric to x-axis.

2. If we replace r with -r, we would get:
-r = -2 + 3*cos(θ )
r = 2 - 3*cos(θ)

NOT same as original, therefore, graph is NOT symmetric to its origin.

3. If we replace θ with -θ and r with -r, we would get:
-r = -2 + 3*cos(-θ )
Since, cos(-θ) = +cosθ, therefore,
r = 2 - 3*cos(3θ)

NOT same as original, therefore, graph is NOT symmetric to y-axis.

Ans: The graph is symmetric to: x-axis only!

19. (Image is attached below) As the question suggests that it is a horizontal ellipse, therefore, the equation for the horizontal ellipse is:


`(x^(2))/(a^(2)) + (y_(2))/(b_(2)) = 1` -- (A)

Since, x = 8f,
y = 18ft,
b = 54ft,

`a^(2)` = ?

Plug-in the values in equation (A),
(A)=>
`(64)/(a^(2)) + (324)/(2916) = 1`

=>
`a^(2)` = 72

Therefore, the equation becomes,
Ans:
`(x^(2))/(72) + (y_(2))/(2916) = 1`


20. Ans: x-axis only
Let's find out!

1. If we replace θ with -θ, we would get:
r = 2*cos(-3θ )
Since, cos(-θ) = +cosθ, therefore,
r = +2*cos(3θ) = Same as original

Therefore, graph is symmetric to x-axis.

2. If we replace r with -r, we would get:
-r = 2*cos(3θ )
r = -2*cos(3θ) = Not same

3. If we replace θ with -θ and r with -r, we would get:
-r = 2*cos(-3θ )
Since, cos(-θ) = +cosθ, therefore,
r = -2*cos(3θ) = Not Same

Ans: The graph is symmetric to: x-axis only!