PLEASE HELP8.08, part 211. 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 = 1B) y squared over 81 minus x squared over 36 = 1C) y squared over 36 minus x squared over 81 = 1D) y squared over 36 minus x squared over 45 = 112. 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 = 1B) y squared over 16 minus x squared over 256 = 1C) y squared over 256 minus x squared over 16 = 1D) y squared over 64 minus x squared over 4 = 113. Eliminate the parameter.x = t - 3, y = t2 + 5 A) y = x2 + 6x + 14B) y = x2 - 14C) y = x2 - 6x - 14D) y = x2 + 1414. Find the rectangular coordinates of the point with the polar coordinates. ordered pair 3 comma 2 pi divided by 3A) ordered pair negative 3 divided by 2 comma 3 square root 3 divided by 2B) ordered pair 3 square root 3 divided by 2 comma negative 3 divided by 2C) ordered pair negative 3 divided by 2 comma 3 divided by 2D) ordered pair 3 divided by 2 comma negative 3 divided by 215. 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 θB) r = 2 + 3 cos θC) r = 2 + 2 cos θD) r = 4 + cos θ18. Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.r = -2 + 3 cos θ A) No symmetryB) y-axis onlyC) x-axis onlyD) Origin only19. 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
Accepted Solution
A:
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 = ± [tex]( \frac{4}{b} ).x[/tex] --- (A) Since asymptotes at y = ± [tex]( \frac{1}{4} ).x[/tex]. --- (B) Compare (A) and (B), you would get, [tex] \frac{4}{b} = \frac{1}{4}[/tex]
=> b=16
The equation (X) would become: [tex] \frac{y^2}{16} - \frac{x^2}{256} = 1 [/tex] (Option-B)
13. Ans: (A) [tex]y = x^{2} + 6x + 14[/tex] Equations given: x = t - 3 --- (equation-1) y = [tex]t^{2}[/tex] + 5 --- (equation-2)
From equation-1, t = x + 3
Put the value of t in (equation-2), [tex]y = (x+3)^{2} + 5[/tex] [tex]y = x^2 + 9 + 6x + 5[/tex] [tex]y = x^2 + 6x + 14[/tex]
Hence, the correct option is (A)
14. Ans: (A)
The polar coordinates given: [tex](3, \frac{2 \pi }{3} )[/tex] = (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 [tex]\frac{2 \pi }{3}[/tex] = 120° => x = [tex]- \frac{3}{2} [/tex]
y = (3) * sin(120°); => y = [tex] \frac{3 \sqrt{3} }{2} [/tex]
Ans: (x,y) = [tex](- \frac{3}{2} ,\frac{3 \sqrt{3} }{2})[/tex] (Option A)
15. Ans: (D) The general forms of finding all the polar coordinates are: 1) When r >= 0(meaning positive): (r, θ + 2n [tex] \pi [/tex]) where, n = integer 2) When r < 0(meaning negative): (-r, θ + (2n+1) [tex] \pi [/tex]) 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) = [tex] \frac{- \pi }{6} [/tex]
When r = +1(r>0): (1, [tex] \frac{- \pi }{6} [/tex] + 2n[tex] \pi [/tex])
When r = -1(r<0): (-1, [tex] \frac{- \pi }{6} [/tex] + (2n+1)[tex] \pi [/tex])
Therefore, the correct option is (D): (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n + 1)π)
(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: